UC-NRLF 


GIFT 


GIFT   OF 

J.  Poster  Plagg 


SYSTEM 


USEFUL   FORMULAE, 


ADAPTED     TO     THE     PRACTICAL     OPERATIONS    OF 


LOCATING  AND   CONSTRUCTING 


RAILROADS: 


A    PAPER    READ    BEFORE    THE    BOSTON    SOCIETY    OF    CIVIL    ENGINEERS, 
DECEMBER,    MDCCCXLIX. 


By    SIMEON     BORDEX, 


CIVIL       E  N  G  I  X  E  E  K. 


BOSTON : 

CHARLES    C.    LITTLE    AND    JAMES    BROWN. 
MDCCCLL 


Entered  according  to  Act  of  Congress,  in  the  year  1850,  by 

CHARLES  C.  LITTLE  AND  JAMES  BROWN, 
in  the  Clerk's  Office  of  the  District  Court  of  the  District  of  Massachusetts 

GIFT 

0 


BOSTON: 

PRINTED   BY   WILLIAM 


PREFACE. 


THE  formulae  contained  in  the  following  pages  were  written,  as  their 
title  indicates,  for  an  original  paper,  which  was  read  before  and  discussed  by 
the  Boston  Society  of  Civil  Engineers. 

The  Society,  having  accumulated  a  number  of  original  papers,  appointed 
a  committee  to  examine  and  report  upon  the  expediency  of  having  them 
published.  After  much  consideration,  the  committee  reported  that,  with 
their  limited  resources  and  present  necessities,  it  would  be  improper  to 
incur  the  expense  of  printing,  but  recommended  an  early  publication  of  this 
paper ;  and  for  that  purpose  the  manuscript  has  been  placed  in  the  hands 
of  the  publisher. 

Notwithstanding  the  obvious  importance  of  constructing  the  curves  of  a 
railroad  upon  the  best  practical  locations,  and  giving  to  their  forms  or 
alinement  the  greatest  degree  of  regularity  practicable,  the  investigation,  or, 
which  is  more  probable,  the  publication  of  anything  like  a  system  of  con 
venient  formulae  to  aid  the  young  engineer,  and  such  others  as  have  not 
had  the  advantage  of  a  good  mathematical  education,  in  the  proper  perform 
ance  of  this  character  of  work,  it  is  believed  has  not  yet  found  a  place 


M166992 


IV  PREFACE. 

upon  the  shelves  of  our  libraries  or  book-stores.  To  supply  this  deficiency 
in  the  library  of  the  civil  engineer,  particularly  the  railroad  engineer,  is 
the  object  of  the  present  paper. 

It  is  not  pretended  that  all  the  formulae  contained  in  this  paper  are 
original.  The  principles  which  have  governed  the  investigations  for  com 
puting  the  elements  required  for  tracing  curves,  where  their  localities  are 
such  as  to  admit  of  the  most  simple  and  convenient  methods,  have,  it  is 
believed,  been  published,  and  are  known  by  most  engineers  who  have  been 
engaged  in  the  construction  of  railroads,  since  the  commencement  of  the 
railroad  system.  Neither  is  it  pretended  that  the  system  of  formulae  is 
complete,  or  that  it  contains  formulae  suited  to  every  case  that  can  arise. 
The  writer  can  only  say,  that  after  considerable  experience  in  the  construc 
tion  of  railroads,  he  does  not  recollect  a  case  presenting  itself  which  would 
not  be  solved  by  some  one  of  the  formulae  ;  and  it  is  believed  that,  with 
slight  modifications,  such  as  any  geometer  would  be  able,  without  difficulty, 
to  make,  they  may  be  adapted  to  all  common  or  ordinary  cases. 

Curves  in  a  railroad,  unless  their  radius  be  very  large,  are  known  to 
be  objectionable ;  but  the  contour  of  the  surface,  the  existence  of  valuable 
buildings,  of  streams,  rivers,  ponds,  oceans,  etc.,  in  the  line  between  the 
points  which  it  is  desired  to  connect,  render  the  adoption  of  curves  neces 
sary.  It  is  likewise  a  well-established  fact,  that  the  greater  the  degree  of 
regularity  and  precision  exercised  in  the  construction  of  curves,  the  more 
safely  and  easily  can  trains  be  run  over  them. 

The  main  objects  of  the  formulae  are  twofold;  viz.,  that  of  enabling  the 
engineer  to  mark  out  the  curves  of  a  railroad  with  the  greatest  degree  of 


PREFACE.  V 

precision  and  convenience,  and  to  locate  them  in  situations  the  most  desir 
able.  To  render  this  subject  clear  and  perspicuous  to  every  one  who  may 
have  occasion  to  locate  or  mark  out  curves,  upon  railroads  and  other  places, 
the  paper  is  commenced  with  the  investigation  of  the  most  simple  problems, 
which  are  succeeded  by  the  more  intricate ;  each  case  being  illustrated 
with  diagrams,  and  accompanied  by  examples  of  computation. 

The  subject  of  switches  and  frogs  being  blended  with  the  elements  of 
turnout  curves,  has  been  considered  in  connection  with  them ;  and  in  their 
arrangement  the  same  objects  have  been  kept  in  view;  and,  for  this  end, 
each  case  has  been  likewise  accompanied  with  a  diagram  and  an  example 
of  computation. 

To  render  the  work  more  useful,  there  have  been  added  formulae  for 
computing  the  cubic  contents  of  excavations  and  embankments,  and  a  formula 
for  computing  the  difference  in  height  to  be  observed  in  laying  down  the 
rails  upon  a  railroad  curve,  based  upon  its  radius  and  the  velocity  of  the 
cars. 

BOSTON,  December,  1850. 


CONTENTS. 


SECTION  PAGE 

1.  —  The  subject  of  connecting  straight  lines  of  different  bearings  by 

simple  curves  considered,  with  an  investigation  of  formula 

for  ascertaining  the  necessary  elements 3 

2. —  Demonstration  of  the  formula  for  angles  of  deflection  for  locating 

a  simple  curve 4 

3.  —  Field  operations   for  locating  or  laying  out  simple   curves  de 

scribed  5 

4.  —  Hints  to  young  engineers 7 

5. — The  subject  of  dividing  curves  into  a  series  or  system  of  long 

chords  considered  and  recommended,  with  practical  hints. . .         7 

6.  —  The  convenience  of  making  the  termination  of  the  long  chords 

correspond  with  even  stations  considered  and  recommended         8 

7.  —  A  method  of  correcting  the  location  of  tangent  points  from  a  re- 

survey  considered  and  explained 9 

8.  —  Formulae  for  ascertaining  the  necessary  elements   for  locating 

simple  curves,  with  examples  of  computation 10 

9.  —  A  method  of  laying  out  simple  curves  by  a  system  or  series  of 

long  chords  described,  with  practical  examples  of  computing 

the  necessary  elements 13 

10.  —  Field  operations  of  laying  out  simple  curves  by  means  of  long 

chords  described 15 

11.  —  The  office  and  field  labors  connected  with  laying  out  simple  curves 

by  the  aid  of  long  chords,  further  considered 17 

12.-— Further  field  operations  of  laying  out  simple  curves  described. .       18 


Vlll  CONTENTS. 

SECTION  PAGE 

13.  —  Practical  operations  and  the  necessary  computations  for  obtaining 

the  elements  of  simple  curves,  and  for  correcting  the  posi 
tion  of  the  tangent  points,  etc.,  based  upon  a  traverse,  con 
sidered  and  illustrated 19 

14.  —  The  same  subject  continued,  with  modifications 25 

15.  —  A  method  of  laying  down  primitive  or  instrument  points  in  a 

curve  by  ordinates  measured  from  the  tangent  lines  continued 

to  apex,  or  the  point  of  their  intersection 34 

16.  —  A  continuation  of  the  same  subject,  with  investigations  of  form 

ulas,  specimens  of  computation,  and  a  description  of  the  field 
work 35 

17.  —  Hints  concerning  the  advantages  of  the  above  named  method 38 

18.  — Reasons  for  not  pursuing  the  subject  of  simple  curves  further  ...       38 

19.  —  Commencement  of  the  subject  of  reversed  curves  —  simplicity  of 

the  problem  for  connecting  two  parallel  lines  by  curves  of 
equal  radii 39 

20.  —  Investigation  of  formulae 39 

21.  —  Specimen  of  computing  elements 42 

22.  — The  same  subject  continued,  with  the  given  elements  varied  —  also 

an  investigation  of  formulae,  with  specimens  of  computation       43 

23  and  24 .  —  The  propriety  of  reversed  curves  with  and  without  a  piece 

of  straight  line  between  them  considered  and  discussed. .  44  to  48 

25.  —  The  same  subject  continued,  and  elucidated  by  diagrams  and  com 

putations  48 

26.  —  The  subject  of  reversed  curves  for  uniting  two  lines  of  different 

bearings  briefly  discussed 51 

27.  —  Investigation  of  formulae  for  obtaining  the  necessary  elements. . .       51 

28.  —  Examples  of  computation 53 

29.  —  A  form  of  reversed  curves  for  uniting  lines  of  different  bearings 

with  one  radius  given  —  also  containing  an  investigation  of 
formulae,  with  specimens  of  computation 55 


CONTENTS.  IX 

SECTION  PAGE 

30.  —  A  form  of  reversed  curves  for  uniting  lines  of  different  bearings, 

having  one  tangent  point,  and  a  centre  angle  to  one  of  the 
curves,  fixed  —  also  an  investigation  of  formulas,  with 
specimens  of  computation 59 

31.  —  A  form  of  reversed  curves  with  or  without  a  piece  of  straight 

line  between  them,  for  uniting  tracks  of  different  or  like 
bearings,  having  the  tangent  points  fixed,  and  the  length  of 
the  radii  governed  by  the  contour  of  the  surface,  or  by  other 
considerations  —  containing  also  an  investigation  of  formulae, 
with  specimens  of  computation G5 

32.  —  The  subject  of  reversed  curves  and  turnout  curves  briefly  noticed      69 

33.  —  Principles  connected  with  the  switch-bar  briefly  discussed 69 

34.  35,  and  36.  —  Remarks  respecting  turnouts  from  a  straight  line  to 

a  side  track  —  with  investigation  of  formulae  and  specimens 
of  computing  the  necessary  elements,  including  the  angle  of 
the  frog 71  to  77 

37.  —  Directions  for  performing  the  field  work  for  laying  out  turnout 

curves 77 

38.  — Investigation  of  formulae  for  ascertaining  the  radius  of  a  turnout 

from  a  straight  track,  suited  to  a  given  frog  ;  with  specimens 

of  computation 77 

39.  40,  and  41. — Turnout  from  the  outside  of  a  curve  in  the  main 

track  —  investigation  of  formulae,  and  specimens  of  compu 
tation,  including  the  angle  of  the  frog ." 81  to  87 

42  and  43.  —  Investigation  of  formulae  for  ascertaining  the  radius  of  a 
turnout  upon  the  outside  of  a  curve  to  suit  a  given  frog  — 
including  examples  of  computation 87  to  93 

44. — Turnout  upon  the  outside  of  a  curve1 — Investigation  of  formulae, 
with  specimens  of  computing  the  necessary  elements,  in 
cluding  the  angle  of  the  frog 93 

45.  —  Investigation  of  formulae  for  determining  a  radius  of  a  turnout 
upon  the  outside  of  a  curve  to  suit  a  given  frog,  including 

examples  of  computation 99 

A 


X  CONTENTS. 

SECTION  PAGE 

46.  —  Discussion  of  the  general  principles  which  combine  in  the  loca 

tion  of  a  turntable  by  the  side  of  the  main  track,  near  a 
water  station,  with  an  investigation  of  formulae  for  determin 
ing  the  elements,  including  specimens  of  computation 103 

47.  —  The  field  work  of  locating  the  turnout  to  the  turntable  described     104 

47.  —  Another  form  of  the  problem  for  locating  a  turnout  to  a  turntable, 
viz.,  by  commencing  at  existing  joints  in  the  rails  of  the 
main  track — with  an  investigation  of  formulae,  specimens 
of  computation,  and  a  description  of  the  field  work 106 

49  and  50.  —  The  subject  of  locating  railroads  with  curves  (or  straight 
lines)  in  places  inaccessible  to  instruments  and  persons,  ex 
cept  in  boats  or  by  some  extraordinary  device,  considered  — 
investigation  of  formulae,  followed  by  specimens  of  com 
putation  —  field  operations  described 109  to  135 

51 .  —  Remarks  on  the  objects  of  this  paper 135 

52.  —  Remarks  on  the  subject  of  vertical  curves  —  formula  for  deter 

mining  their  elements  —  examples  of  computation 137 

53.  —  Hints  respecting  the  management  of  the  elements  obtained 146 

54.  —  Subject  of  vertical  curves,  continued — formula   deduced,  and 

practical  operations  described 146 

55.  —  Same  subject,  continued 150 

56.  —  Remarks  upon  the  subject  of  bending  rails  for  curves  of  short 

radii  —  formula  for  obtaining  the  elements  necessary  to  mark 

out  a  pattern  board  —  examples  of  computation 153 

57.  —  Several  formulae  for  determining  the  radii  of  curvature  in  railroad 

tracks,  with  examples  of  computation 158 

58.  —  Remarks  upon  the  elevation  of  the  outside  rail  of  a  railroad 

curve,  with  formulae  and  specimens  of  computation 161 

59  to  62.  —  Demonstrations  of  formulae  for  determining  the  cubic  con 
tents  of  pyramids,  wedges,  parallelepipeds,  etc 166  to  174 

63  to  66.  —  Formulae  for  determining  the  cubic  contents  of  excavations 

and  embankments,  with  examples  of  computation. . . .    174  to  188 


E  R  B  A  T  A. 


On  page  77,  ninth  and  tenth  lines  from  top,  for  "  =    90°  —  Sw 
+  \  (180°  —  C)  "  read  "  (90°  +  Sw)  —  i  (180°  —  C)." 

On  page  87,  seventh  line  from  bottom,  for  "=  r  -f~  \  h  -f-  d" 
read  "  r  —  %  h  +  d," 

On  page  113,  second  line  from  top,  for  "S  75°  08'  35"'07  E" 
"875°  08'  35"  -37  E." 


On  page  119,  fourteenth  line  from  bottom,  for  "We  then  dis 
cover  "  read  "  We  then  determine,"  etc. 


AN    INVESTIGATION 


0  F 


USEFUL    FORMULAE, 


ADAPTED       TO 


THE   PRACTICAL  OPERATIONS  OF  LOCATING  AND  LAYING  OUT 
RAILROAD   CURVES. 


In  3mie0tigatioE  of 


ETC.      ETC.      ETC. 


( 1 )  COMMENCING  with  the  most  simple  operations  and  forms 
of  computations,  we  shall  assume  as  the  most  simple  of  railroad 
curves,  such  as  will  unite  two  direct  or  straight  lines  of  different 
hearings  in  such  a  manner  that  each  of  these  lines  will  "become 
tangents  to  said  curve.  Having  run  these  straight  or  tangent 
lines  to  their  intersections  at  A,  (which  intersection  we  shall 
hereafter  term  their  apex,  see  Fig.  1,)  and  determined  the  magnitude 
of  their  angle,  we  then  proceed  to  determine  the  most  favorahle 
location  for  the  track.  In  order  to  do  this,  we  examine  the  contour 
of  the  surface,  and  select  the  position  we  think  the  most  favorahle 
opposite  the  angle  at  A ;  then  we  run  and  measure  from  A  to  this 
most  favorahle  place  for  the  track  hefore  selected,  in  a  direction 
that  will  hisect  the  angle  A ;  that  is,  from  A  to  Ss,  and  this 
distance  we  represent  "by  b.  We  then  proceed  to  ascertain  a  radius 
for  the  curve,  which  shall  pass  through  the  point  83  and  so  run  into 
the  straight  lines  that  they  shall  be  tangents  to  said  curve. 


4  USEFUL     FORMULAE. 

Putting  r  for  the  radius  in  feet,  or  the  unit  of  measure ; 

"       C  for  the  angle  at  the  centre  of  the  curve  corresponding 

with  the  subtense  of  its  arc  =  (180°  —  A ;) 
"       t  for  the  distance  from  apex  to  the  points  of  commence 
ments  of  the  curve,  or  where  the  direct  line  becomes  a 
tangent  to  said  curve,  viz.,  from  A  to  T,  (See  Fig. ;) 
We  have  sin.  J  C  :  b  : :  sin.  (J  C  +  i  A)  :  t  =  * sin'^  +  * A) 
and  cos.  J  A  :t  ::  sin.  \  A  :  r  ==^AA^^^+A^== 
tan.  \  K.I  cot.  J  C.  (1) 

Having  thus  determined  the  radius  of  the  desired  curve  for 
uniting  the  aforesaid  tangent  lines,  and  the  distance  from  their 
apex  to  their  tangent  points,  or  points  of  commencement  of  said 
curve,  we  will  now  proceed  to  investigate  some  of  the  most  simple 
and  practical  methods  of  locating  or  laying  out  the  same. 

(  2  )  We  consider  first  what  we  shall  call  the  method  of  de 
flections.  To  explain  this  operation,  let  us  suppose  the  arc  or 
curve  to  be  divided  into  such  equal  parts  as  a  chord  of  the  length 
of  the  chain  contemplated  to  be  used  will  span.  We  have  in  the 
course  of  our  practice  generally  used  for  this  purpose  a  chain  fifty 
feet  in  length ;  as,  by  using  a  short  chain,  the  chords  and  the  arcs 
(if  the  radius  be  of  much  magnitude)  are  nearly  of  the  same  length, 
which  affords  a  great  convenience  in  determining  the  deflecting 
angle  corresponding  to  such  short  chords  (consisting  of  fractions 
of  the  chain)  as  it  will  frequently  be  found  desirable  to  use  at  the 
commencement  and  termination  of  curves,  that  we  may  be  enabled 
to  keep  up  a  continuous  notation  of  equi-distant  stations. 

Many  engineers,  1  am  aware,  use  a  chain  of  100  feet,  which, 


DEFLECTING      ANGLES.  5 

if  the  radius  be  not  of  considerable  magnitude,  will  give  a 
perceptible  difference  between  the  length  of  the  chords  and  the 
corresponding  arcs  they  span.0 

To  determine  the  angle  of  deflection  corresponding  to  the  length 
of  the  chain  to  be  used,  we  represent  the  length  of  the  chain  by  cJi ; 
the  angle  of  deflection  by  D ;  and  the  angle  at  the  centre  of  the 
curve,  corresponding  to  its  subtense,  by  C'. 

Supposing  T,  Si,=  the  chord  ch,  we  have  in  the  triangle  C  T  Si, 

\  (180°—  C')  =  the  angle  T  =  Si  ; 
Now   as   the   angle  A  T  C  =  a  right  angle,  or   90°,  we   have 

90"— (J«*=*-)  =  D; 

Multiplying  by  2,  180°  — (180°— C')  =  2D  ; 
Subtracting  180°,  and  changing  signs,  we  have 

2D  =  C'; 

Consequently,  D=  i  C'  ;  (2) 

Bisecting  ch  we  have  r  :  K  : :  J  cli  :  sin.  \  C'=  ^~  =  sin.  D       (3) 

It  sometimes  happens  that  we  wish  to  know  the  value  of  D  at 

the  commencement  of  our  computations ;  expanding  the  foregoing, 

sin.  D  =  * ch  cot-  *,A  tan"  *  c  (4) 

(  3  )  Having  thus  determined  the  angle  of  deflection,  we  now  are 
prepared  for  locating  or  marking  out  the  curve. 

Adjusting  a  good  theodolite  to  the  tangent  point  T,  with  its 
principal  telescope  (or  the  telescope  by  which  angles  are  determined) 

*  In  delineating  a  curve  by  the  method  of  deflections,  it  will  be  inconvenient  to  make  the 
stations  further  apart  than  the  length  of  the  chain  used. 


6  USEFUL     FORMULAE. 

pointing  in  the  direction  of  A,  and  with  its  watch  telescope  (all 
theodolites  should  be  provided  with  watch  telescopes)  pointing  to 
any  convenient  well-defined  mark,  lay  off  the  angle  D,  and  then 
stretching  the  chain  from  T,  fix  its  terminus  Si  in  range  with  the 
principal  telescope,  the  point  thus  marked  will  he  in  the  curve ; 
then,  laying  off  another  angle  of  deflection  which  will  read  upon  the 
instrument  =  2D,  stretch  the  chain  from  Si  to  82,  fixing  the  ter 
minus  at  82  in  the  range  indicated  by  the  main  telescope ;  the 
points  thus  formed  will  also  be  in  the  curve.  And  in  like  manner 
we  proceed  to  fix  the  points  Ss,  84,  85,  etc.,  until  the  curve  is  com 
pleted  ;  or,  as  it  more  frequently  happens,  as  long  as  the  contour 
of  the  surface  will  permit  us  to  sec  distinctly. 

Let  us  now  suppose  an  obstacle  which  will  prevent  seeing  beyond 
83.  We  then  remove  our  theodolite  to  that  station ;  and,  after 
having  duly  adjusted  it,  with  its  main  telescope  pointing  at  the 
station  thus  left,  and  the  watch  telescope  to  some  convenient  mark, 
lay  off  an  angle  equal  to  180°,  minus  the  sum  of  the  deflections 
made  at  the  first  station,  plus  one  deflection.  Then,  stretching  the 
chain  from  the  station  where  the  theodolite  is  now  adjusted,  place 
the  other  terminus  in  range  with  the  principal  telescope,  as  here 
tofore  described.  Then,  proceed  in  the  manner  above  described,  to 
lay  off  deflections  and  chords,  until  you  connect  with  the  straight  or 
tangent  lines,  or  as  long  as  the  contour  of  the  country  will  admit, 
when  the  instrument  must  be  again  changed,  and  the  like  operations 
performed  until  the  whole  curve  is  completed.  If  the  curve  is  not 
measured  by  whole  chords  =  c/t,  the  deflection  for  the  fraction  of  ch 
will  be  to  that  of  ch  in  the  proportion  the  fraction  bears  to  a  whole ; 
which,  although  not  strictly  correct,  is  sufficiently  near  for  practice. 
\Ve  here  remark  that  we  do  not  recommend  that  more  than  10  or 


DIVIDING     CURVES,  7 

12  chords  of  50  feet  each  should  be  laid  down  with  the  theodolite 
at  one  station,  it  being  conducive  to  accuracy  not  to  permit  a  great 
difference  between  the  direction  of  the  chord  (cji)  and  the  direction 
of  the  pointing  of  the  telescope  of  the  instrument. 

(  4  )  We  have  thus  endeavored  to  describe  a  practical  method  of 
laying  out  one  of  the  simplest  of  railroad  curves  which  shall  unite 
two  straight  lines  having  different  bearings ;  but,  as  there  is  a 
great  variety  of  methods  to  accomplish  this  end,  which  may  be 
resorted  to,  some  of  which  seem  to  possess  peculiar  adaptation  to 
certain  localities,  I  have  thought  it  would  not  be  uninteresting  to 
describe  some  of  them,  believing  a  few  hints  of  this  kind  would  lead 
the  new  beginner  to  different  modes  of  reasoning  and  investigation ; 
and,  if  he  possesses  a  tolerable  knowledge  of  the  elements  of  plain 
geometry,  he  will  be  able  always  to  select,  if  not  the  method  best 
adapted  to  the  circumstances  of  the  case,  at  least  one  well  suited 
and  convenient. 

(  5 )  If  the  curve  be  of  large  radius,  (and  curves  cannot  well  be 
constructed  with  too  large  a  radius,)  and  its  location  suits  the 
contour  or  surface  of  the  country,  and  the  apex  angle  be  large,  a 
very  convenient  and  accurate  method  of  proceeding  will  be  to  divide 
the  curve  into  a  series  of  segments  of  some  500  or  600  feet  each, 
as  may  be  thought  best ;  taking  care  as  far  as  possible  to  make  the 
terminus  of  each  segment  an  even  station ;  °  it  will,  however,  fre 
quently  happen  that  the  number  of  the  stations  marking  the  tangent 

*  In  the  location  and  construction  of  a  railroad,  it  has  been  found  convenient  in  practice  to 
divide  the  centre  line  into  equal  parts,  technically  called  stations,  which  contain  a  given  number 
of  the  units  of  measure  used  in  the  construction.  In  the  United  States,  the  foot  has  been  taken 
for  the  unit  of  measure,  and  the  railroad  stations  one  hundred  feet  asunder. 


8  USEFUL     FORMULAE. 

points  will  contain  a  fraction,  and  of  course  in  this  case,  the  segments 
should  contain,  besides  a  number  of  whole  chords,  the  fractional 
chord  the  case  requires. 

Having  determined  upon  these  preliminary  considerations,  we 
first  calculate  the  relative  position  of  points  on  the  line  between 
the  tangent  points  and  apex,  wrhich  shall  correspond  to  radii  of  the 
curve  passing  through  the  termini  of  the  several  segments ;  also 
the  distance  from  the  points  upon  the  radii  of  the  curve,  together 
with  the  angles  the  radii  made  with  the  tangent  lines ;  the  points 
thus  computed  are  readily  determined  and  marked  with  a  good 
degree  of  accuracy,  and  become  instrument  stations,  from  whence 
the  intermediate  stations  are  readily  filled  up  as  before  described, 
by  chords  and  deflections,  with  less  liability  to  practical  errors 
than  the  preceding  method. 

( 6  )  If  the  curve  be  of  large  radius  and  the  apex  angle  small, 
the  distance  of  the  tangent  points  from  the  apex  and  from  the 
tangent  lines  to  the  curve,  will  become  too  great  to  be  measured 
conveniently  with  a  proper  degree  of  accuracy  ;  under  this  condition 
of  the  case,  we  divide  the  curve  into  a  convenient  number  of  seg 
ments,  taking  care  as  before  to  have  their  termini  to  correspond 
with  the  even  stations. 

Having  thus  determined  upon  the  divisions,  we  compute  the 
chords  and  angles  corresponding  therewith,  and  then  proceed  to  lay 
off  the  angles  with  great  care,  and  measure  the  chords  with  as 
great  a  degree  of  accuracy  as  is  practicable,  carefully  marking 
the  termini  of  each  chord.  If  the  chords  and  angles  when  laid 
down  correspond  with  the  tangent  points  and  tangent  lines,  we 


TESTING      PRIMITIVE      CHORDS.  9 

proceed  to  put  in  the  intermediate  stations  by  adjusting  the 
theodolite  to  the  terminus  of  one  of  these  chords ;  then,  pointing  to 
the  other  terminus,  we  proceed  to  lay  down  the  intermediate  chords 
and  stations  by  the  aid  of  deflections,  in  the  manner  described  in 
the  foregoing.  It  is  usual  to  fill  up  the  intermediate  stations 
formed  by  two  of  the  primitive  chords  without  changing  the  instru 
ment  ;  having  done  this,  we  move  our  instrument  to  another 
terminus  common  to  two  chords,  which  have  not  been  filled  up  with 
the  intermediate  stations,  and  in  like  manner  we  proceed  until  we 
complete  the  curve. 

(  7  )  If,  in  running  our  first  chords  until  we  have  exhausted  our 
computations,  we  do  not  find  the  work  to  correspond  with  a  proper 
degree  of  accuracy  to  the  tangent  points,  and  tangent  lines,  we 
make  a  connection  with  the  tangent  points,  and  carefully  ascertain 
the  length  of  the  chord,  and  the  magnitude  of  the  angle  with  the 
tangent  line ;  and,  with  the  elements  thus  obtained,  we  recompute 
the  work,  and  determine  the  relative  position  of  the  tangent 
points,  and  fix  them  anew.  We  then  proceed  to  perform  the  work 
of  laying  down  the  primitive  chords  a  second  time ;  when,  if  there 
have  been  no  mistakes  made,  the  work  will  prove  practically 
accurate.  It  is  very  seldom  that  a  second  computation  will  be 
needed  until  the  road  is  graded,  when  the  measurement  taken  upon 
the  graded  surface  will  not  be  likely  to  be  identical  with  the 
primitive  measurement;  then,  a  resurvey,  and  a  re-establishment 
of  the  tangent  points,  become,  if  not  absolutely  necessary,  desir 
able  ;  but,  if  the  first  survey  was  performed  with  any  tolerable 
degree  of  care,  and  the  grading  well  finished,  no  difficulty  will  be 
experienced  in  laying  the  curve  upon  the  graduated  road-bed, 
without  changing  its  radius. 
C 


10  USEFUL      FORMULA. 

(  8  )  Having  thus  briefly  described  the  principles  which  govern 
us  in  laying  down  simple  curves,  we  would  now  introduce  an 
example,  accompanied  by  specimens  of  calculation  for  each  par 
ticular  case. 

PRACTICAL      EXAMPLES      OF    CALCULATIONS     SUITED      TO      THE 
CASES      DESCRIBED      IN      THE      FOREGOING      PAGES. 

Assuming  A  =  160°,  then  will  C  =  180°  —  A  =  20° 
b  =  132  feet 
ch  =  50  feet 

To  find  the  deflection,  we  have  (4)  sin.  D  =  — 

Thus,  |  A  =  80°  00'  00"         cot.  =  9-2463188 

i  C   =    5°  00'  00"         tan.  =  8-9419518 

£  ch  =    25  feet  log.  =  1-3979400 

6        =  132  feet  co.  ar.  log.  =  7-8794261 


D       =    0°10'02".64      sin.  =  7-4656367 

Iii  this  case  it  will  be  seen  that  D  =  0°  10'  02". 64  is  an  incon 
venient  angle  to  add  or  subtract,  or  even  read  upon  the  instrument ; 
we  therefore,  to  remedy  this  objection,  adopt  D  =  0°  10',  which 
will  not  materially  change  the  length  of  the  radius  or  the  location 
of  the  curve.  This  change  requires  that  we  base  our  calculations 
for  determining  the  elements  needed,  upon  D  =  0°  10'  ;  therefore, 
to  find  the  radius  we  have 

Sin.D:l(*::R:r  =  -A^D  (5) 

and  to  find  t  we  have 

Cos.  J  C  :  r  : :  sin.  J  C  :  t  =  tan.  J  C .  r  (G) 

(representing  by  t  the  distance  from  the  apex  to  the  tangent  point, 
or  from  A  to  T  on  the  diagram.) 


COMPUTATION      OF      ELEMENTS.  11 

Thus,     £  ch  ==  25  feet  -  -  -  -     log.  =  1  -3979400 

D      =    0°  10'  co.  ar. sin.  =  2-5362745 


Consequently,    r       =  8594-38  feet log.  =  3-9342145 

Then,     \  C  =  10°  00'  00"        tan.  =  9-2463188 


t        =1515-42  feet log.  =  3-1805333 

Having  thus  ascertained  the  radius  r  =  8594-38  feet,  and  the 
distance  from  apex  to  tangent  point,  (t=  1515 '42  feet,)  we  now 
proceed  to  find  b.  By  analogy  we  have 

Sin.  (i  C  +  i  A)  :  t  ::  sin.i  C  :  b  =  »$££*  =  t  tan.  J  C  (7) 

Then,    t       =  as  above,  •  •  •  •     log.  =  3  - 1805333 

£  C  =  5°  00'  00"  tan.  =  8-9419518 

b       =  132-58  feet          log.  =  2-1224851 

At  commencement ) 

>  b  =  132-00  feet 
we  assumed         ) 

Difference, =      0-58  feet 

Thus  it  appears  that  the  change  of  the  radius  to  cause  it  to 
correspond  with  D  =  0°  10'  will  only  change  the  location  of  the 
curve  from  the  position  designed  for  it  0' 58  feet;  a  quantity  too 
small  to  be  generally  accounted  anything  in  choosing  the  position 
of  a  curve. 

We  now  proceed  to  find  the  length  of  the  curve.  Putting  r"  for 
an  arc  in  seconds  =  radius,  and  C"  for  the  number  of  seconds 
contained  in  the  centre  angle  which  measures  the  curve,  and  a  the 
arc  subtending  the  angle  C",  we  have  by  analogy 

r"  :  r  : :  C"  :  a  =  ^  (8) 

Thus,        r    =  8594.38  feet  log.  =  3.9342145 

C"  =  7200"  log.  =  4.8573325 

r"  =  co.  ar.         log.  =  4.6855749 

a     —  3000.00  feet  log.  =  3.4771219 


COMPUTING      ELEMENTS.  13 

Now,  that  we  may  show  this  matter  as  complicated  as  it  is 
generally  found  in  practice,  let  us  suppose  the  pin  at  T  marking 
the  first  tangent  point,  to  be  numbered  140 '38;  the  integer  repre 
senting  the  whole  number  of  the  stations,  and  the  decimals  the 
fractional  part  of  the  space  beyond  station  140.  We  have  found 
the  length  of  the  arc  a  equal  30  whole  stations ;  which,  added 
to  140.38,  gives  170*38  for  the  number  of  the  other  tangent 
station  T'. 

( 9 )  Let  us  now  proceed  to  show  the  method  of  laying  out  the 
curve  by  the  method  of  a  series  of  long  chords  corresponding  to 
about  750  feet  of  the  arc,  which  we  afterwards  fill  up  by  simple 
deflections  and  50  feet  chords  =  ch. 

In  the  first  place  we  have  supposed  the  first  tangent  pin  to  bear 
the  number  140*38  ;  if  we  now  add  7*12  stations,  it  will  make  the 
number  147.50;  therefore  our  first  chord  will  extend  from  station 
140*38  to  station  147*50.  We  now  add  7*  50  stations  to  station 
147*50,  which  makes  the  number  155 ;  therefore  our  second  chord 
extends  from  station  147*50  to  station  155.  We  then  add  7*50 
stations  to  station  155,  which  increases  the  number  to  162*50; 
therefore  our  third  chord  extends  from  station  155  to  162*50.  We 
now  add  7*88  stations  to  162*50,  which  increases  the  number  to 
170*38,  and  brings  us  to  the  tangent  point  T'. 

Having  assumed  ch  =  50  feet,  (which  is  practically  the  same 
length  of  the  arc  it  spans  when  the  radius  is  of  considerable 
length,)  we  must  now  proceed  to  determine  the  length  of  the 
several  long  chords  we  have  divided  the  arc  into.  (See  the  diagram 
on  the  opposite  page.) 


14  USEFUL      FORMULAE. 

Our  first  long  chord  corresponds  to  712  feet  of  arc.  We  How 
propose  to  find  the  centre  angle  which  it  subtends  by  the  following 
formula. 

Eepresenting  the  arc  by  a  and  the  centre  angle  in  seconds  by  C", 
we  have 

r  :  r"  : :  a  :  C"  =  -^  (9) 

and  further  we  have 

Cos.  \  C"  :  r  : :  sin.  C"  :    Ch?=  ~~^lf  (10) 

Thus,      r    =  8594-38  feet  co.  ar.        log.  =  6-0657855 

a    =    712-00  feet  log.  =  2-8524800 

r"  =  log.  =  5-3144251 

C"  =   17088"  log.  =  4-2326906 

C"  reduced  to  degrees  and  minutes  =  4°  44'  48"  sin.  =  8-9177692 

£  C"     "  "  "        =  2°  22'  24"      ar.   co.      cos.  =  0-0003719 

r    log.  =  3-9342145 

First  chord  =  Ch  =  711  -  79  feet     =  log.  =  2  -  8523556 

The  arc  corresponding  to  the  succeeding  chord  is  750  feet  long, 
and  is  composed  of  15  whole  deflections,  and  each  deflection  being 
10',  the  centre  angle  will  be  equal  to  twice  that  number  of  deflec 
tions,  or  300',  which  amounts  to  5°  00'  00". 

Now,  by  formula  (10)  we  have 

We  distinguish  the  primitive  or  long  chords  by  Cfi  in  contradistinction  with  the  50  feet  or 
deflecting  chords,  which  are  represented  by  ch. 


DETERMINATIONS      OF  ANGLES.  15 

C"      in  degrees  =  5°  00'  00"  sin.  =  8-9402960 

AC"  in  degrees  =  2°  30'  00"  co.  ar.  cos.  =  0-0004135 

r=     log.  =  3-9342145 


Second  chord  =  749-763  feet        log.  =  2-8749240 

Third  chord,  of  course,  will  be  of  the  same  length,  viz.,  749  '763. 

The  arc  corresponding  to  the  fourth  chord  is  788  feet  long,  and 
we  find  C"  and  Oh  by  formula  (9)  and  (10) 

Thus,     r     co.  ar.    log.  =  6-0657855 

a    =  788  feet  log-  =  2-8965262 

r" log.  =  5-3144251 


C"  =  18912"  log.  =  4-2767368 

C"    reduced  to  degrees  and  minutes  =  5°  15'  12"  sin.  =  8-9617037 

i  C"       "        "        "         "        "        =  2°  37' 36"  co. ar. cos.  =  0-0004554 
r    =  log.  ==  3-9342145 


Fourth  chord  =  Ch  =  787-723  feet  log.  =  2-8963736 

We  have  not  attempted  to  make  the  foregoing  calculations 
strictly  exact ;  our  angles  being  always  taken  to  correspond  with 
the  nearest  second,  which  in  most  cases  gives  a  greater  degree  of 
accuracy  than  we  can  practically  execute. 

(10)  Having  thus  computed  the  centre  angles  and  chords 
corresponding  with  the  proposed  division  of  the  arc,  we  will 
endeavor  to  give  the  method  of  laying  out  the  work. 

In  the  first  place  we  will  determine  the  angles  at  the  tangent 
points,  and  the  several  stations  which  are  to  mark  the  termini  of 
the  divisions  of  chords. 


16  USEFUL      FOKMUL^. 

Commencing  at  the  tangent  point  numbered  140*38,  our  first 
operation  is  to  determine  the  angles  in  the  several  triangles,  viz. ; 
T  C  1,  1  C  2,  2  C  3,  3  C  T.  (See  Fig.  2.) 

The  angle  C",  which  we  have  determined,  will  correspond  to  C, 
and  in  triangle  T  C  1,  C  =  4°  44'  48",  (see  section  9,)  each  of 
these  triangles  being  isosceles,  and  having  computed  their  centre 
angles,  we  have  only  to  deduct  them  from  180°,  and  half  the 
remainder  will  be  the  value  of  each  of  the  remaining  angles.  Now, 
calling  triangle  TCI  No.  1,  and  1  C  2  No.  2,  and  2  C  3  No.  3, 
and  3  C  T  No.  4,  we  proceed  to  find  their  angles  in  the  order  we 
have  named  them. 

DETERMINATION    OF    THE    ANGLE    AT    FIRST   TANGENT   POINT,    OR    T. 

The  angle  at  centre  of  curve  for  triangle  No.  1=4°  44'  48" 
and  180°-42044/48"  =  87°  37'  36"  =  angle  at  T,  or  at  sta 
tion  140-38  and  147-50. 
"     angle  at  centre  of  curve  for  triangle  No.  2  =  5°  00'  00" 


2 

and  155. 

"  angle  at  centre  of  curve  for  triangle  No.  3  =  5°  00'  00" 
and  i8o° -50  oo' GO-  =  87o  3(y  =  angle  at  stations  155  and 

162-50. 

"  angle  at  centre  of  curve  for  triangle  No.  4  =  5°  15'  12" 
and  180°-52015'12"  =  87°  22'  24"  =  angle  at  stations 
162-50  and  170-38. 

Having  thus  prepared  the  angles  for  the  several  stations  above 
named,  and  for  the  purpose  of  rendering  our  description  easier  to 
be  understood,  we  arrange  them  in  their  order,  as  in  the  foregoing 
diagram. 


MEASURING      ANGLES. 


17 


(11)  Having  thus  represented  our  work  as  shown  in  the  dia 
gram,  we  now  proceed  to  ascertain  the  angles  at  the  termini  of 
the  several  chords.  Thus, 


At  T,  or  station,  ----   140-38  =  90°  -f  87°  37'  36"        =  177°  37'  36" 

"1,          "            ----  147-50  =  87°  37'  36"  -f-  87°  30'=  175°  07'  36" 

"2,          "             ----   155-00  =  87°  30'  +  87°  30'          =  175°  00'  00" 

"3,          "             ----   162-50  =  87°  30'  -f-  87°  22'  24"=  174°  52'  24" 

"   T',       "            ----   179-38  =  87°  22'  24"  +  90          =  177°  22'  24" 

Having  ascertained  the  angles  for  each  station,  which,  for  con 
venience,  we  write  upon  the  several  radii  connecting  said  stations 
with  the  centre  of  the  curve  in  the  diagram,  we  proceed  to  compute 
the  length  of  the  several  chords  which  span  their  respective  arcs  ; 
commencing  with  triangle  TCI,  which  is  called  No.  1,  and  pursuing 
the  calculations  in  the  order  shown  in  the  diagram.  Thus, 


T    =  87°  37'  36"  •  •  • .  co.  ar. 
r     =  8594-38  feet 
C    =     4°  44'  48" 

Ch  =  711-79  feet  ••• 


•  •  sin.  =  0-0003727 
log.  =  3-9342145 
sin.  =  8-9177692 


0 


=  log.  =  2-8523564 

sin.  =  0-0004135 

log.  =  3-9342145 
sin.  =  8-9402960 
=  log.  =  2-8749240 

3     =87°  22'  24" co.  ar.   ••••  sin.  =  0-0004565 

r     —  log.  =  3-9342145 


QQ 

co 

1      =  87°  30' 

00" 

....  co 

3 

T3 

0 

a 

T       = 

fe 

£VJ 

C    =     5°  00' 

00" 

1 

i 

Ch  = 

749 

76  feet 

C    =  5°  15'  12" 

Ch  =  787 -72  feet 


sin.  =  8-9617037 


=  log.  =  2-8963747 

(12)  Having  thus  prepared  our  work,  we  proceed  to  adjust  the 


18  USEFUL      FORMULAE. 

theodolite  to  station  T,  or  according  to  the  locating  stations,  to 
No.  140*38,  with  its  principal  telescope  pointing  in  the  direction  of 
the  line  of  the  road;  which,  it  is  presumed,  has  been  properly 
marked.  Then,  laying  off  an  angle  of  177°  37'  36",  we  measure 
in  the  direction  indicated  by  the  telescope,  711*79  feet  to  1  or  to 
station  147-50;  then,  moving  the  instrument  to  147*50,  and 
pointing  the  principal  telescope  to  T,  we  lay  off  an  angle  of 
175°  07'  36",  and  measure  in  the  direction  indicated  749*76  feet 
to  2  or  station  155.  Then,  moving  the  instrument  to  station  2  and 
pointing  at  1,  we  lay  off  an  angle  of  175°,  and  measure  in  the 
direction  indicated  749*76  feet  to  3  or  to  station  162*50.  Then, 
moving  the  instrument  to  3  and  pointing  at  2,  we  lay  off  an  angle 
of  174°  52'  24",  and  measure  in  the  direction  indicated  787*72  feet 
to  T  or  to  station  170*38  ;  which,  if  our  angles  and  measures  bring 
us  direct  to  T  or  near  to  it,  we  presume  the  work  to  be  correctly 
done.  We  should,  however,  before  pronouncing  the  work  correct, 
place  our  instrument  at  T',  and  pointing  the  telescope  to  3,  lay  off 
the  angle  with  the  line  of  the  road,  and  if  this  agrees  with  the 
computed  angle,  I  think  we  may  then,  without  hesitation,  pronounce 
the  work  correct. 

But,  if  our  angles  and  measures  do  not  bring  us  direct  to  T'  or 
near  by  it,  we  then  point  our  telescope  to  T',  and  ascertain  the 
angle  indicated  by  the  instrument,  and  measure  the  distances  as 
correctly  as  we  can,  which  we  duly  note  down  in  our  field  book.  We 
then  move  to  T'  with  our  instrument,  and  pointing  its  telescope 
to  3,  we  measure  the  angle  with  the  line  of  the  road,  which  we  also 
note  in  our  field  book. 

With   the   data   thus   obtained,   we   proceed   to   recompute   the 


CURVE      FROM      SUPPOSED      FIELD      NOTES.  19 

elements  of  a  curve  that  will  unite  the  two  lines  without  materially 
varying  the  location  of  the  track  from  the  points  which  we  have 
just  fixed. 

(13)  In  order  to  show  practically  the  performance  of  these 
operations,  we  will  make  up  the  following  as  the  field  notes  of  a 
survey  for  locating  the  curve  ahove  described. 

Commencing  at  T  with  instrument  pointing  in  the  direction  of 
the  road, 

we  laid  off  the  angle  177°  37'  36"  and  measured  711-79  feet  to  station  1 

At  station  1    we  laid  off  the  angle  175°  07'  36"  and  measured  749-76  feet  to  station  2 

"           2              "              "  175°  007  00"  "             749-76               "  3 

"           3              "              «  175°  00'  00"  "             751-51               "  T 

"          T              "              «  177°  14'  48"  in  the  direction  of  road. 


Sum,    ....   880°  GO7  00" 

Having  obtained  the  field  notes  of  our  traverse,  our  first  operation 
will  be  to  deduce  from  them  the  angle  at  apex,  and  at  the  centre  of 
the  curve.  We  here  remark  that  the  sum  of  the  angles  at  apex  and 
at  the  centre  of  the  curve,  always  amount  to  180°,  and  of  course  one 
must  be  a  supplement  to  the  other. 

To  ascertain  either  of  the  angles,  viz.,  at  the  apex,  or  at  the 
centre,  a  variety  of  formulae  might  be  deduced,  but  it  is  presumed 
the  following  is  as  convenient  as  any ;  viz.,  subtract  the  sum  of  all 
the  angles  from  as  many  times  180°  as  there  are  angles,  and  the 
remainder  will  be  the  angle  at  the  centre  of  the  curve. 

It  will  bo  seen  that  we  have  noted  in  our  field  book  five  angles, 
whose  sum  amounts  to  880° ;  now  5  X  180  =  900 ;  and  900  —  880 


20  USEFUL      FORMULAE. 

=  20°  ==  the  angle  at  the  centre  of  the  curve,  which  compares  with 
the  angle  we  had  formerly  ascertained. 

Our  next  step  is  to  ascertain  the  relative  position  of  the  present 
points  T  and  T'  with  respect  to  A,  and  also  the  position  they  should 
occupy  to  suit  the  radius  we  have  heretofore  deduced. 

We  know  of  no  more  convenient  method  of  determining  this 
problem,  than  "by  working  up  the  traverse,  (as  the  ship  captains 
call  it,)  and  for  that  purpose  we  will  assume  the  line  of  the  road 
extended  from  T  to  apex,  whatever  may  he  its  direction,  as  hearing 
due  north,  and  predicate  the  bearings  of  the  other  lines  upon  it,  as 
indicated  by  the  angles.  Thus, 

Angle  at  T  =  177°  37'  36"  and  180°  —  177°  37'  36"  leaves  20  22'  24"  N.  W.  to  1 

"  "  1  =  1750  07'  36"  =  3520  45'  12"  and  (2  X  180  o)  —  352  °  45'  12"  =    7O  14'  48"        "         2 

"  "  2  =  1750  oo'  00"  =  527°  45'  12"    "    (3  X  180 o)  —  527 o  45'  12"  =  12°  14'  48"        "         3 

"  "  3  =  1750  00'  00"  =  7020  45>  12"    «    (4  X  180°)  —  702  o  45'  12"  =  17°  14'  48"        "       T' 

"  «  T'  =  1770  14'  48"  =  8800  00'  00"    "    (5  X  180°)  —  880  o  00'  00"  =  20  o  00'  00"        " 

being  the  direction  of  the  road. 

Computing  the  northings  and  westings  of  the  foregoing  traverse, 
we  have 

No.  1.    N.  W.  2°  22'  24"     sin.  =  8-6171119  cos.  =  9-9996273 

711  -79  feet    log.  =  2-8523458  log.  =  2-8523458 


29-475    "  log.  =  1-4694577  711  -17  feet  log.  =  2-8519731 

No.  2.     N.  W.  7°  14'  48"  sin.  =  9-1008572  cos.  =  9-9965171 

749- 76  feet  log.  =  2-8749223  iog.  =  2-8749223 

94-576    "  log.  =  1-9757795  743- 77  feet  log.  =  2-8714394 


WORKING      THE      TRAVERSE. 


21 


No.  3.     N.  W.  12°  14'  48"     sin.  =  9-3265833 
749-76  feet    log.  =  2 •  8749223 


cos.  =  9-9900028 
log.  =  2-8749223 


159-04     "      log.  =  2-2015056     732-70  feet  log.  =  2-8649251 


No.  4.    N.  W.  17°  14'  48"     sin.  =  9-4720042 
751-51  feet    log.  =  2-8759348 


COS.  =  9-9800203 
log.  =  2-8759348 


222-814     "    log.  =  2-3479390     717-72  feet  log.  =  2-8559551 

Then,  summing  up  the   computed   northings   and  westings,  we 
have  as  follows : 


BEARING. 

DISTANCE. 

NORTHING. 

WESTING. 

From  T  to  No.  1  == 

N.  W.  2  22  24 

711-79 

711-17 

29-475 

«       i      »       2  = 

"       7  14  48 

749-76 

743-77 

94-576 

"       2      "        3  = 

"     12  14  48 

749-76 

732-70 

159-040 

"       3      "        4  = 

"     17  14  48 

751-51 

717-72 

222-814 

Total   — 

2905-36 

505-905 

Having  summed  up  the  traverse,  w£  now  proceed  to  find  the 


bearing  and  distance  from  T  to  T'. 


Putting  N  =  the  sum  of  the  northings,  and  calling  it  the  cos. ; 
W  =        "  "       westings,  "  "       sin. ; 

C   =  the  bearing  sought ; 
D  =  the  distance  from  T  to  T'. 


We  then  have  by  analogy  the  following  formulae  : 


N  :  W  : :  ft  :  tan.  C  = 


\v 


\\ 


and  sin.   C  :  W  : :  R  :   I)        =  slf-e 
or  cos.      C  :  N  : :  R  :  D        =  — N-c- 


(A) 
(B) 
(C) 


[Fio.     3.] 


TESTING      COMPUTATIONS.  23 

Performing  the  computations  indicated,  we  have 

W=  505-905        log.  =  2-7040690 

N  =2905-36    co.  ar.  log.  =  6-5368001 

C  =  9°  52'  40"       tan.  =  9-2408691 

C  =  9°  52'  40"  co.  ar. sin.  =  0-7656168 

W  =  log.  =  2-  7040690 

D  =  2949-08  feet     log.  =  3-4696858 

We  now  have  in  the  triangle  ATT',  the  side  T  T'  and  the  data  for 
finding  the  unknown  angles.*  Then,  to  find  the  distances  A  T  and 
A  T'  we  have 


sin.  A  :  D  : :  sin.  T  :  A  T'  =5 
and  sin.  A  :  D  : :  sin.  T'  :  A  T  == 


P  sin.  T 
sin.  A 

P  sin.  T' 
sin.  A 


To  prevent  confusion  in  our  diagram,  or  to  render  our  work  more 
plain,  we  reconstruct  the  figure  of  the  triangle  ATT'.  (See  figure 
on  preceding  page.) 

To  perform  the  computations  indicated,  we  have 

A  =160°  00'  00"  co.  ar.    sin.   =  0-4659483 

D  =  2949-08  feet  log.  =  3-4696861 

T'  =  10°  07'  20"  sin.   =  9-2448918 

A  to  T  =  1515-396  feet  log.  =  3-1805262 

Sum  of  logs,  of  A  and  D  =  log.  =  3-9356344 

T  =9°  52'  40"  sin.  =  9  •  2343832 


A  to  T'  =  1479-7  feet  =  log.  =  3-1700176 

*  The  data  for  finding  the  unknown  angles  are  the  relative  bearings  of  the  sides  of  the  triangles,  as 
found  in  the  preceding  computations. 


24  USEFUL      FORMULAE. 

Having  thus  computed  the  distance  from  apex  to  the  points  T 
and  T',  we  will  now  ascertain  the  length  those  lines  should  be  to 
suit  the  contemplated  radius  ;  thus  we  have 

Cos.  i  C  :  r  ::  sin.  J  C  :  *  =  tan.  J  C.r 

r       =  8594- 38  feet    log.  =  3-9342145 

1  C  =  10°  00>  00"     tan.  =  9  •  2463188 

t         —  1515- 42  feet     log.  =  3-1805333 

t  in  the  present  case,  as  in  our  former  notation,  represents  the  dis 
tance  required  from  A  to  T  and  also  from  A  to  T',  to  suit  the  con 
templated  curve,  which  in  the  present  instance,  has  a  radius  of 
8594-38  feet,  and  an  apex  angle  A  =  1GO°. 

It  appears  from  the  above  computations  that  T  should  be 
moved  from  the  apex  1515-42  --  1515-396  —  -024  feet,  which 
amount  in  practice  is  so  small  that  we  should  consider  T  correctly 
located,  and  doubtless  our  calculations  would  have  given  its  location 
exact,  had  we  been  careful  in  the  management  of  the  fractions ; 
but  it  is  not  so  with  T.  The  computations  show  that  T'  should  be 
moved  from  apex  1515-42  -  -  1479-17  =  3G-25  feet.  After 
having  moved  T'  from  apex  3G-25  feet,  in  the  direction  of  the 
line  of  the  road,  there  will  be  no  doubt,  if  the  previous  traverse 
upon  which  the  calculations  have  been  based,  has  been  correctly 
measured,  that  the  contemplated  curve  could  be  accurately  located 
between  the  tangent  points  as  corrected. 

The  above  calculations  have  been  based  upon  a  traverse  consist 
ing  of  chords  of  an  arc  correctly  laid  down,  with  the  exception  of 
the  last  course  ;  but,  had  the  traverse  been  a  random  one,  the  results 
arrived  at  would  have  been  equally  exact,  with  only  this  dift<erence, 


FIELD      NOTES      OF      A      TRAVERSE.  25 

that  the  points  T  and  T  would  not  probably  have  been  found  so 
near  their  proper  location. 

The  practice  explained  above  will  be  found  useful  in  locating 
curves  by  the  side  of  rivers,  ponds,  oceans,  mountains,  rough  sur 
faces,  etc. ;  in  short,  wherever  it  is  found  inconvenient  to  run  the 
direct  lines  to  their  intersections,  or  to  apex,  and  to  measure  there 
from  to  the  points  T  and  T'.  In  running  traverses  for  the  purpose 
of  obtaining  the  elements  for  the  location  of  curves  under  the  con 
ditions  suggested  above,  it  will  be  found  convenient,  if  not  abso 
lutely  necessary,  that  some  portion  of  the  traverse  should  be  so 
made  as  to  give  the  relative  position  of  such  points  as  the  contour 
of  the  surface  or  other  considerations  may  render  it  desirable  that 
the  location  should  pass  through.  We  now  proceed  to  give  a  prac 
tical  example. 

(14)  The  following  practical  example  supposes  the  direct  or 
straight  lines  to  be  united  by  the  curve,  to  have  been  located  and 
marked  by  some  convenient  device,  and  the  angles  given  below  to 
have  been  measured  by  a  common  theodolite,  and  the  lines  by  a 
chain,  thus :  ° 

Field  notes  of  a  traverse  for  obtaining  the  elements  of  a  curve 
for  uniting  the  lines  T  and  T'. 

Commencing  at  station  0,  corresponding  in  the  diagram   to  T, 


*  This  remark  would  seem  to  be  uncalled  for,  as  it  can  make  no  difference  in  the  computation  how 
the  lines  and  angles  are  obtained  5  but  it  frequently  happens  that  the  traverse  is  obtained  by  means  of 
a  triangulation,  which  would  sometimes  present  the  matter  in  a  different  form.    The  writer  has  had  a 
number  of  cases  that  could  not  have  been  well  performed  in  any  other  manner. 
E 


26  USEFUL      FOKMULJE. 

with   telescope  pointing  in  the  direction  of   the  located  line,  we 
measured  as  follows,  viz. : 


C  The  curve  should  pass  through  or 
Station  0  =  169  29  45  =  1200  feet  to  station  1  J     near  this  point,  which  of  course 

(      governs  the  radius. 
"        1  =  170  00  15  =     900     "  "          2 

"        2  =  175  04  30  =     750     "  "          3 

«        3  =  1G4  40  10  =  1525     "  "          4  corresponding  in  diagram  to  T' 

"       4  =  140  45  20  =  in  the  direction  of  the  located  line  of  the  road. 


820  00  00 
180  X   5    =  900  00  00 


80  =  angle  at  centre  of  curve. 


Assuming  the  line  from  T  to  A  as  hearing  due  north,  whatever 
its  course  may  he,  we  deduce  the  following  relative  courses  for  the 
several  lines  of  the  traverse,  and  hy  formula  (A)  ascertain  the 
relative  hearing  from  T  to  T'. 


STATION 

BEARINGS. 

DISTANCES. 

NORTHINGS. 

WESTINGS. 

0  to  1  = 

N.  W.  10°  30'  15" 

1200 

1179-890 

218-7685 

1   "  2  = 

"       20°  30'  00" 

900 

843-005 

315-1866 

2  "  3  = 

"       25°  25'  30" 

750 

677-361 

321-9970 

3  "  4  = 

"        40°  45'  20" 

1525 

1163-197 

995-5700 

3863-453       1851-5221      log.  =  3-2675289 
3863-453       log.  =  3-5869756 

Relative  bearing  from  T  to  T'  =  N.  W.  25°  36'  20"  -22    tan.  =  9  •  6805533 

Having  thus  obtained  the  hearing  from  T  to  T',  we  now  proceed 
to  compute  the  distance ;  hy  formula  (B)  and  (C)  we  have 
Sin.  C  :  W  : :  R  :  D  =  ~r    or 
Cos.  C  :  N  ::  R  :  D  =  c^ 


DISTANCE      TANGENT      TO      APEX. 


27 


In  these  analogies,  C  represents  the  course  from  T  to  T  ;  W  the 
westing ;  N  the  northing ;  D  the  distance. 

We  frequently,  as  we  shall  in  the  present  instance,  use  "both 
formulae,  for  the  purpose  of  proof. 


W  =  1851-5221  log.  =  3-2675289 

C    =  25°  36'  20".22  sin.  =  9-6356588 
D    =  4284-2  feet        log.  =  3-6318701 


N  =  3863-453  log.  =  3-5869756 

C   =  25°  36'  20".22  cos.  =  9-9551055 


D  =  4284-2  feet        log.  =  3-6318701 


Our  next  step  in  practice  is  to  ascertain  the  distances  from  T  and 
T  respectively  to  the  point  where  the  direct  lines  would  intersect ; 
or,  in  other  words,  to  the  apex. 

We  have  ascertained  the  angle  of  the  centre  of  the  curve  to  he 
80°.  Of  course  the  angle  at  apex  will  he  100°.  The  hearings 
which  we  have  ascertained  also  indicate  the  angles  ;  thus,  in  the 
imaginary  triangles  we  are  about  to  solve,  we  have  supposed  the 
line  from  T  to  A  to  hear  due  north.  Then,  by  computation,  the 
line  from  T  to  T'  bears  N.  W.  25°  36'  20". 22,  which  gives  the 
angle  at  T  the  same  number  of  degrees  as  the  bearing.  From  the 
traverse  or  the  table  of  angles,  in  our  field  notes,  we  deduce  the 
bearing  of  the  located  line  from  T  to  be  N.  W.  80°  00'  00". 

These  bearings  indicate  tlie  following  angles,  viz.,      at  A   =  100°  00'  00". 00 

As  before  stated,  at  T    =     25°  36'  20".22 

T'  =     54°  23'  39".78 

Proof,          Sum  =  180°  00'  00".00 

With  these  angles,  and  the  distance  from  T  to  T  =  1),  the  dis 
tances  T  A  and  T'  A  are  readily  found  ;  thus, 


[Fia.     4.] 


MEASUREMENT   TANGENT   TO   APEX.        29 

sin.  A  :  D  : :  sin.  T'  :  A  T 


(10) 
and  sin.  A  :   D   : :   sin.  T  :   A  T'  l 


A        =  100°  00'  00"     ar.  co.     sin.   =  0-0066485 

D       =  4284-2  feet  log.  =  3-6318701 

T        =  54°  23'  39".78  ski.   =  9-9101139 


A  T  =  3536-98  feet  log.  =  3-5486325 

Again,    A        =  100°  00'  00"     co.  ar.  sin.  =  0.0066485 

D        =  4284-2  feet  log.  =  3-6318701 

T         =  25°  36'  20".22  sin.   =  9-6356588 

A  T'  =  1880-085  feet  log.  =  3-2741774 

For  the  triangle  A  T\  (see  diagram)  we  have  by  our  measurement 
and  computations  the  sides  A  T  =  3536-98  feet,  and  1  T  =  1200 
feet  with  their  included  angle  =  10°  30'  15"  to  find  A  1  and  the 
unknown  angles. 

For  convenience  in  the  enunciation  of  the  formula,  let  A  T  =  a, 
1  T  =  b,  C  =  the  given  angle,  and  A  and  B  the  unknown  angles  ; 
A  representing  the  unknown  angle  opposite  the  side  a,  and  B  the 
unknown  angle  opposite  the  side  b.  We  then  have  a  -f-  b  .* 
a  ^  b  : :  tan.  }  (180°—  C)  :  tan.  J  (A  ^  B)  and  \  (180°  —  C)  + 
\  (A  co  B)  =  the  angle  opposite  the  longest  side,  and  J  (180°  — 
C)  —  i  (A  oo  B)  =  the  angle  opposite  the  shortest  side. 

In  the  calculations  which  follow  we  change  the  symbols  from 
those  given  in  the  formula,  so  as  to  have  them  conform  to  the 
letters  and  figures  given  upon  the  diagram. 


30 


USEFUL      FORMULAE. 


A  T 

3536-98      Given  angle  T 

180°  00'  00" 
=             100  30'  15" 

Irp 

2)  1690  29'  45/' 

JL                               •  —  • 

1200 

84  O  44'  52"  5 

JQO-    —  6'3°44984 

DifF                  — 

]O(y   —  3-3686550 

i   (180°      T)  — 

tan   —  1«0365792 

£   (A  co  1)         = 

79°  26'  43".  14 

tan.  =  0-7297256 

Z_  at  1             = 
l_  at  A           = 
l_  at  T 

164°  11'  35"-  64 
5°  l&  09"-  36 
10°  30'  15".  00 
180°  00'  00".  00 

Z_  at  1 
AT 
T 
A  1 


164°  11'  35". 64  co.  ar. 
3536-98  feet 
10°  30'  15" 
2367-21982  feet 


sin.  =0-5648026 
log.  =  3-5486326 
sin.  =9-2608034 


log.  =  3-3742386 


A 

Tl 

T 


Proof, 


=       5°  18'  09" -36  co. 
=  1200  feet 
=     10°  30'  15" 
=  2367-21982  feet 


...  &in.  =  1  •  0342539 
. ..  log.  =  3-0791812 
...  sin.  =9-2608034 


,.  ..  ]0jr  =3-3742385 


Having  obtained  all  the  elements  of  the  triangle  A  1  T  we  rep 
resent  the  side  A  1  as  found  above  in  feet,  or  in  general  in  the  unit 
of  measure  by  bm,  and  as  we  find  it  in  proportion  to  the  radius  of 
the  curve  =  unity,  by  b. 

As  it  becomes  convenient  generally  to  use  the  letters  standing 
against  the  angles  in  each  triangle,  and  as  some  of  them  are  com 
mon  at  least  to  three  triangles,  it  becomes  necessary  to  occasionally 
accent  some  of  them,  that  we  may  understand  their  different  values 


DETERMINATION      OP      RADIUS.  31 

in  the  investigation.  We  have  throughout  our  investigations  rep 
resented  the  apex  angle  by  A.  We  shall  continue  to  give  to  A  that 
value  in  the  following  investigation,  with  the  exception  of  angle  A 
in  the  triangle  AIT  (which  in  general  will  not  be  a  multiple  of 
the  apex  angle)  we  shall  therefore  represent  it  by  A'. 

Commencing  our  investigation  with  the  triangle  A  T"  C,  and 
representing  T"  C  the  radius  of  the  curve,  by  unity,  and  the  radius 
of  the  tables  by  E,  we  have 

Sin.iA:  1  ::  B:  AC=-1BJnr  (11) 

In  the  triangle  A  C  1  we  have  01  =  the  radius  of  the  curve  =  1 ; 
therefore,  representing  the  angle  at  1  by  G,  we  have 

1   .  sin.  (i  A  —  A')   : :  AC  :  sin.  G  =   ""1*7*°  (12) 

For  convenience  representing  the  line  A  C  by  6?,  we  have 

Sin.  G  :  d  ::  sin.  [(J  A  —  A')  +  G]  :  b  =  rf  sin'[(^n~GAO+GL  (13) 
then  b  :  bm  ::  1  :  r=  >-  (14) 

where  r  represents  the  radius  of  the  curve  in  the  unit  of  measure. 

Or,  probably  the  following  formula  would  be  rather  more  simple 
for  calculation  than  the  above,  (13)  viz.: 

Sin.  (i  A  --  A')  :  1  ::  sin.  [(i  A  -  -  A')  +  G]  :  b  =    , 

sin.  f(*  A  —  A')  +  H]  / 1  o'  \ 

sin.  i  A- A'  ViJ  ) 

then,  as  above  b  :  bm  : :    1   :  r  =  —  *—  (14') 

where  r  represents  the  radius  of  the  curve  in  the  unit  of  measure. 

Example  of  calculation,  formula  (12)  sin.  G  =  J 


32  USEFUL      FORMULA. 

We  found  A  by  the  reduction  of  the  foregoing  traverse  (see  pre 
ceding  pages)  =  100°,  also  A'  (noted  as  A)  in  the  triangle 
A  1  T  =  5°  18'  09". 3G.  Therefore, 

£  A  —  A'  =  50°  —  5°  18'  09".3G  =  44°  41'  50". 64      sin.  =  9-8471791 
A  50°  00'  00".00  co.  ar.  sin.  =  0-1157460 


G     (ambiguous)  66°  39'  38"  .22  sin.  =  9-9G29251 

G     (corrected)  113°  20'  21".78 

Having  found  G,  we  proceed  to  find  the  radius  =  r.     By  formula 
(13')  and  (14')  we  have 


7)  -  -  _«5i_(       ^ 


A'  +*  • 


sin.  (£  A—  A)  b  ~    sin.  (J  A—  A'  -f  G) 


£  A  —  A'  =       44°  41'  50"-  64 

G  =     113°  20'  21".  78 


A  — A'-fG=     158°  02'  12". 42         co.  ar.     sin.  =  0-4271153 

A  —  A'  =       44°  41'  50". 64         ....  ....     sin.  =  9-8471791 

=  2367-2198  feet  log.  =  3-3752386 


r          =  4462-035  feet  log.  =  3-6495330 

We  thus  find  the  radius  of  the  curve,  or  r  =  4462'  035  feet. 
The  deflection  for  a  chord  =  50  feet  will  be  (3)  sin.  D  =   -i^- 

Thus, ^  ch  =  25  feet  log.  =  1-3979400 

r        =  4462-035  feet        co.  ar.        log.  =  6-3504670 
D       =0°  19'  15"- 67  sin.  =  7-7484070 

But,  as  0°  19'  15"  -07  makes  an  inconvenient  number  to  add  or 
subtract,  we  choose  for  the  angle  of  deflection  (D)  =  0°  19'  15", 
and  adopt  a  radius  which  shall  agree  therewith.  This  change  in 
the  radius  will  not  materially  alter  the  location  of  the  curve. 


POSITION      OF      TANGENT      POINTS. 

To  find  a  radius  corresponding  to  a  deflection  of  0°  19'  15",  we 
(5)  r=  4^ 

i  ch  =  25  feet  log.  =  1-3979400 

D       =  0°  19'  15"     co.  ar.     sin.   =  2-2518454 


r         =  4464-63  feet  log    =  3-6497854 

We  have  thus  ascertained  the  radius  of  a  curve  which  will  cor 
respond  to  the  location  selected.  It  now  remains  to  ascertain  the 
tangent  points,  or  points  of  commencement  and  end. 

We  have  (6)  representing  the  whole  centre  angle  hy  C, 
t  =  tan.  J  C .  r 

Thus,      .»••     £  C  =  40°  00'  00"     tan.  =  9-9238135 

r        =  4464-63  feet    log.  =  3  6497854 

t         =  3746-27  feet    log.  =  3-5735989 

We  found  (page  29)  A  T  ==  3536-98  feet,  and  A  T=  1880-085 
feet ;  and,  t  being  374G  •  27  feet,  we  have 

t          =  3746-27 
AT    =  3536-98 


209-29  feet,  the  distance  T  should  be  moved  from  A 

Again,    t        =  3746-27 
AT'  =  1880-085 


1866-185  feet,  the  distance  T'  should  be  moved  from  A 

Having  'moved  the  points  T  and  T'  to  their  positions  indicated 
above,  and  marked  in  the  diagram  T'"  and  T",  the  curve  may  be 
laid  out  and  marked  by  any  method  the  engineer  might  think  best 
suited  to  the  locality. 


34  USEFUL      FORMULAE. 

(15)  We  will  add  one  more  method  of  locating  simple  curves, 
principally  applicable  to  large  apex  angles,  and  which  may  in  some 
instances  be  practised  beneficially  with  apex  angles  somewhat  acute, 
provided  the  radii  be  not  of  great  length. 

This  method  supposes  that  the  locality  will  admit  of  the  lines  to 
be  connected  by  the  curve,  to  be  run  or  extended  to  their  intersec 
tion,  so  that  their  apex  angle  may  be  measured  directly  ;  and  that 
the  contour  of  the  surface  is  such  that  measurements  may  be  taken 
with  a  good  degree  of  accuracy,  from  the  apex  to  the  tangent  points 
along  these  extended  lines,  and  from  these  extended  lines  to  the 
location  of  the  curve. 

Having  premised  thus  much,  let  us  suppose  these  extended  lines, 
which  we  shall  hereafter  call  the  tangent  lines,  intersect  each  other 
at  an  angle  of  170°,  and  it  be  desirable  to  connect  these  lines  by  a 
curve  corresponding  to  a  deflecting  angle  of  6'  for  a  chord  of  50 
feet.  The  elements  required  to  be  obtained  from  computation  are, 

First,  Kadius  of  the  curve. 
Second,  Length  of      do. 
Third,  Distance  from  apex  to  tangent  points. 
Fourth,  The  number  of  primitive  points  convenience  requires  to 
be  fixed  in  the  curve. 

To  ascertain  the  radius,  we  have  (5' )  r  =  -^fjj- 

£  ch  =  25  log.  =  1-3979400 

D   «  0°  6'  00" co.  ar.  sin.  =  2-7581229 

Radius  =  r        —     14323-95  feet  locr-  =  4-1560629 


PRIMITIVE      POINTS      IN      CURVES.  35 

To  find  the  length  of  the  curve,  we  have  (8)  a  =   r  '£" 

r   =  log.  =  4-1560629 

180°  —  170°  =  C"  =  36,000" log.  =  4-5563025 

r     co.  ar.          log.  =  4-6855749 


Length  of  arc  =  a    =  2500-02  feet  log.  =  3-3979403 

To  find  the  distance  from  apex  to  tangent  points,  (G)  t  =  tan.  J  C  .  r 

£  C  =    5°  00'  00"  tan.  =  8-9419518 

r     log.  =  4-1560629 

Apex  to  tan.  =  t       =    1253-18  feet  log.  =  3-0980147 

We  have  found  the  length  of  arc  =  2,500  feet ;  if  we  now  sup 
pose  T  to  hear  an  even  number  in  the  locating  stations,  say  540,  we 
may  divide  the  arc  into  five  equal  parts  of  five  hundred  feet  each, 
which  will  cause  every  point  of  division  to  fall  on  an  even  station  in 
the  location.  This  division,  of  course,  divides  the  centre  angle  into 
five  equal  parts ;  and,  as  C  =  180  —  170  =  10°,  the  centre  angle 
corresponding  to  each  division  will  be  -°—  «=  2°  00'  00". 

(16)  Having  determined  to  divide  the  curve  into  five  equal 
parts,  we  now  compute  the  distances  from  each  of  these  dividing 
points  in  the  curve  to  the  tangent  lines,  in  the  direction  of  the  radii 
passing  through  them.  (See  Fig.  5.) 

Denoting  the  tangent  points  by  T  and  T' ;  and  the  divisions  of  the 
curve  by  1,  2,  3,  4  ;  and  the  corresponding  divisions  of  the  tangent 
lines  by  a  b  c  d;  and  representing  by  Ci  the  centre  angle  corre 
sponding  to  the  arc  T  1,  and  by  €2  the  centre  angle  corresponding 
to  the  arc  T  2.  (The  arcs  T'  4  and  T'  3,  being  similar  to  T  1  and 
T  2,  will  not  need  separate  expressions.)  In  the  general  investiga- 


[FiQ.     5.] 


PRIMITIVE      POINTS      IN      CURVES.  37 

tions,  we  however  denote  the  centre  angles  by  C,  and  we  have  this 
analogy, 

Cos.  C  :  r  : :  sin.  C  :  T  a  =  tan.  C  .  r  (15) 

and  Cos.  C  :  r  : :  R  :  C  a        =  -^^ 

and  we  have  -— -^ r  =  a,  1  =  etc.  (1C)  corresponding  with  the 

centre  angle.     Now,  substituting  Ci  and  C2  for  C,   as  explained 
above,  we  have 

Ci     =  2°  00'  00"  tan.  =  8  •  5430838 

r        =  14323-95  feet          log.  =  4-1560629 

T'  to  d  and  T  to  a=      500-20  feet          log.  =  2-6991467 

Ci      =  2°  00'  00"  co.  ar.   cos.  =  0-0002646 

r        =14323-95  log.  =  4-1560629 

C  to  d  and  C  to  a  =  14332-69  log.  =  4-1563275 

.*.  d  to  4  and  a  to  1  =          8-74  feet 

Again,  Cz     =  4°  00'  00"  tan.  =  8-8446437 

r        =  log.  =  4-1560629 


T'  to  c  and  T  to  6  =     1001  •  65  feet  log.  =  3  •  0007066 

C2  =  4°  00'  00"  co.  ar.  cos.  =  0-0010592 

r        =  14323-95  feet  log.  =  4-1560629 

C  to  c  and  C  to  b  =  14358-93  log.  =  4-1571221 

.'.  c  to  3  and  6  to  2   =        34-98  feet 


Having  thus  ascertained  all  the  elements  necessary  to  this 
peculiar  method,  we  may  now  measure  from  T  to  a,  in  the  direction 
T  A=  500*2  feet;  and  the  same  distance  from  T'  to  cZ,  in  the 
direction  T'  A  ;  and  then,  with  the  theodolite  at  a,  and  pointing  to 
T,  lay  off  the  complement  angle  of  Ci,  and  measure  8 '74  feet  to  1, 
fora  point  in  the  curve  corresponding  to  station  540 '00  -f-  5 '00 
=  545  of  the  location.  Then,  remove  the  theodolite  to  d,  and 


38  USEFUL      FORMULAE. 

pointing  at  T,  lay  off  the  complement  angle  of  Ci,  and  measure  in 
the  direction  indicated  8  •  74  feet  to  4,  for  a  point  in  the  curve  cor 
responding  to  station  (540-00  +  20'  00)  =  560  of  the  location. 
We  now  measure  from  d  (1001-63  —  500-20)  =  501  -43  to  c,  and 
the  same  distance  from  a  to  b ;  then,  with  the  theodolite  at  6,  and 
pointing  at  T,  lay  off  the  complement  angle  of  €2,  and  measure  in 
the  direction  indicated  24-98  feet  to  2,  for  a  point  in  the  curve  cor 
responding  to  station  (540-00+  10*00)  =550  of  the  location. 
Then,  moving  with  the  theodolite  to  c,  and  pointing  to  T ,  lay  off  the 
complement  angle  of  Ca,  and  measure  in  the  direction  indicated  24-98 
to  3,  for  a  point  in  the  curve  corresponding  to  station  (540*00  -f- 
15-00)  =  555  of  the  location. 

(17)  We  now  take  the  theodolite  successively  to  each  of  the 
points  we  have  established  in  the  curve ;    and,  by  deflections  and 
corresponding  chords,  complete  the  work. 

This  method  of  laying  out  curves  is  found  exceedingly  convenient 
in  woodlands,  as  not  being  so  liable  to  mistakes  which  might  lead 
the  location  astray  as  other  methods,  and  will  frequently  save  much 
trouble  in  chopping  timber. 

We  are  aware  that  the  example  we  have  given  in  the  foregoing 
is  one  of  the  most  convenient  that  the  problem  admits  ;  but  we  think 
the  principle  will  be  sufficiently  comprehended  to  apply  it  readily 
and  without  difficulty  in  its  most  complicated  form,  without  further 
explanation. 

(18)  We  have  now  completed  all  we  contemplated  respecting 
the  investigations  of  the  practical  operations  of  laying  out  simple 


KEVERSE      CURVES.  39 

railroad  curves.  It  was  not  our  purpose  to  pursue  these  investiga 
tions  until  the  subject  was  exhausted.  That  would  have  taken  a 
long  time,  and  might  have  occupied  much  room,  and  it  is  more  than 
probable  that  it  would  have  exceeded  our  ingenuity  and  ability. 
But  we  hope  we  have  said  enough  to  give  the  proper  direction  to  the 
inquiries  of  the  young  or  inexperienced  engineer,  and  to  convince 
him  of  the  necessity  of  making  the  study  of  the  elements  of 
geometry  and  trigonometry  (if  it  be  proper  to  make  the  distinction, 
or  to  class  them  under  different  heads)  a  matter  of  the  first 
importance. 

(19)  We  now  proceed  to  the  consideration  of  reverse  curves  ; 
the  most  simple  form  of  which  is  to  unite  two  parallel  lines,  which, 
continued,  will  not  intersect  or  run  into  each  other,  by  curves  of 
equal  radii. 

This  problem  is  of  such  simplicity  as  to  admit  of  many  forms  of 
construction,  and  great  variety  of  formulae ;  and,  had  we  not  come  to 
the  conclusion  to  give  a  formula  for  laying  out  curves  to  unite  rail 
road  lines  under  every  condition  which  has  occurred  in  our  practice, 
we  think  we  should  have  passed  by  this  problem  without  considering 
its  properties.  We  therefore  content  ourselves  with  giving  the  fol 
lowing  rules  and  formulae. 

(20)  Let  T  A  and  T"  B  represent  two  lines  of  railroad  having 
the  same  bearing,  but  so  located  that  they  will  not  intersect  by  ex 
tension. 

Let  T  C  and  T'  C'  =  the  radii  which  we  suppose  to  be  equal  in 
length,  and  which  in  the  investigation  we  shall  represent  by  r. 


[Fia.     6.] 


REVERSE      CURVES.  41 

Draw  the  line  T  C  and  T  C'  at  right  angles  with  T  A  and  T  B, 
and  make  each  =  r ;  then,  drawing  a  line  from  C  to  C'  will  inter 
sect  the  curves  at  their  reversing  point  X,  (see  figure,)  so  also  a  line 
drawn  from  T  to  T  will  intersect  the  curves  at  the  same  point, 
and  will  also  bisect  the  line  C  C'. 

We  now  extend  the  line  B  T'  until  it  intersects  the  line  or  radius 
T  C  at  E,  and  as  T  C  is  drawn  at  right  angles  with  T  A,  and  T  A 
and  B  T'  having  the  same  bearing,  it  is  obvious  that  the  angle  at 
the  intersection  will  be  a  right  angle.  Then,  having  measured  T'  II 
and  T  R,  or  ascertained  their  length  by  computation,  we  have  by 
considering  the  line  =  T  T  as  radius,  and  the  line  T  R  as  a 
cosine,  and  the  line  R  T  as  a  sine  of  the  angle  T  T'  R,  this  analogy 
to  find  the  angle  T'.  Representing  the  sine  by  «,  the  cosine  by 
c,  the  radius  by  R,  then  will 

c  :  s  ::  R  :  tan.  T  =  -f-  (17) 

By  letting  fall  a  perpendicular  upon  the  line  T  X  from  C,  we 
divide  the  triangle  T  C  X  into  two  equal  right-angled  triangles, 
viz.,  C  M  T  and  C  M  X  ;  now,  as  the  triangles  T  T'  R  and  T  C  M 
have  the  angle  T  common  to  both,  it  is  obvious  they  are  similar ; 
the  angles  T'  and  C  must  therefore  be  equal,  and  in  the  triangle 
T  C  X  the  angle  C  will  be  equal  to  twice  T';  then,  resolving  the 
triangle  T  R  T,  we  find  the  length  T  T,  half  of  which  is  equal  to 
T  X  ;  then  solving  the  triangle  C  X  T,  we  find  the  side  T  C  = 
radius  =  r.  To  execute  these  computations  by  the  aid  of  the  trig 
onometrical  functions,  we  have  in  the  triangle  T  T'  R  to  find 
J  T  T'  which  we  represent  by  a. 

Sin.  T   :  *   ::  R  :   T  T';  wherefore  -T2T1     =  a  =  =  -3^-^- 
or,  Cos.  T  :  e  ::  11  :  T  T';  */'-  =  a  =-c~<,7.r 

G 


42  USEFUL      FOKMUL.E. 

Then,  in  the  triangle  T  C  X  we  have 

Sin.  C  :  a  ::  cos.  T'   •  r  =  ^^  =  ^-^     (18) 

Having  thus  found  the  radius,  and  the  centre  angle  C  of  the  one 
half  the  curve,  it  is  evident  that  the  other  half  must  contain  iden 
tical  elements. 

(21)  The  curve  may  be  laid  out  in  accordance  with  such  of  the 
formulae  described  in  the  foregoing  pages,  relating  to  simple  curves, 
as  the  engineer  may  think  best  suited  to  the  condition  of  things 
and  the  contour  of  the  surface. 

For  the  purpose  of  presenting  an  example  of  computation,  we 
will  suppose 

T'  E  =  c  =  12GO  feet 

E  T    =  s  =     150    " 
Then,  by  (17)  Tan.  T'  =  ~ 

Wherefores     =     150  feet  log.  =  2-1760913 

c     =1260     "      co.  ar.    log.  =  6-8996295 


T'  =  6°  47'  20". 31    tan.  =  9-0757208 

C  being  equal  to  2  T,  we  have  (18)  r  =    sin  cc   2 
Wherefore  T'  =    6°  47'  20". 31 


C  =  13°  34'  40"- 63  CO.  ar.  sin.  =  0-6293609 

2  "     "  log.  =  9  •  6989700 

c     =  1260  log.  =  3  -1003705 

r  =2683-47  log.  =  3-4287014 


REVERSE      CURVES.  43 

(22)  We  will  now  vary  the  given  elements,  by  supposing  the 
radius  =  r  =  2683 '47  feet,  and  the  distance  between  the  centres 
of  the  tracks  =  s  =  150,  to  find  the  relative  positions  of  T'  and 
T.  This  case  supposes  a  beginning  point  to  have  been  selected. 
Suppose  that  point  to  be  at  T.  Then,  from  T  we  draw  the  radius 
T  C  =  r  at  right  angles  to  the  line  T  A,  and  from  T  we  extend 
the  line  T  C  or  r  in  the  opposite  direction,  a  distance  =  r  —  (or 
minus)  the  distance  between  the  tracks  (which  we  have  supposed  to 
be  150  feet)  to  D ;  and  from  D  draw  the  line  D  C'  at  right  angles 
with  the  line  D  C  ;  then,  with  the  line  C  C' ,  equal  in  length  to  2  r, 
intersect  the  line  D  C'  at  C'  ;  then  will  D  C'  =  T  E,  representing 
T'  E  as  in  our  former  proposition  by  c ;  and  then,  solving  the  tri 
angle  C  C'  D,  we  have  by  trigonometry,0 

2  r  :  E  : :  (2  r  —  s)  :   sin.  C'  =  ^-f^  (19) 

and     Sin.  C'   :   (2  r  —  s)  : :  cos.  C'   :   c  =  cot.  C'  (2  r  —  s)    (20) 

The  complement  of  the  angle  C'  found  by  the  above  formula  = 
the  angles  T  C  C  ==  T'  C'  C. 

EXAMPLE     OF     COMPUTATION. 

By  formula  (19) 

2r  =  5367  co.  ar.        log.  =  6-2702685 

(2  r  —  s)  =  5217         log.  =  3-7174298 

C'       =  76°  25'  19".50  sin.  =  9-9876893 

By  formula  (20) 

C'       ==  76°  25'  19".50  cot.  =  9-3829486 

(2  r  —  s)  =  log.  =  3-7174208 

c  =  1260  nearly         log.  =  3-1003694 

*  These  formula  are  generally  based  on  trigonometry,  as  being  more  convenient  or  better  suited 
to  logarithms. 


44  USEFUL      FOKMUL^E. 

The  complements  to  C'  as  found  above  =  90  —  76°  25'  19". 5  = 
13°  34'  40". 5. 

(23)  There  being  a  difference  of  opinion  among  practical  engi 
neers  respecting  the  propriety  of  reversing  curves  without  the 
intervention  of  a  piece  of  straight  track  between  them,  we  have 
thought  it  would  not  be  improper,  before  we  proceed  further  with 
our  investigations  of  formulae  for  reverse  curves,  to  add  the  follow 
ing  discussion  of  the  causes  of  the  lateral  shocks  experienced  in 
railroad  cars  when  entering  upon  a  curve ;  and  also  the  necessity 
for  a  piece  of  straight  track  between  reversing  curves. 

From  any  investigations  that  we  have  been  enabled  to  make,  we 
cannot  discover  any  reason  why  a  car  should  receive  a  lateral  shock 
when  running  off  a  straight  track  on  to  a  curve,  provided  the  dis 
tance  between  the  flanges  of  the  car  wheels  and  the  distance 
•between  the  insides  of  the  rails  are  the  same  ;  and  the  outside  rail 
of  the  curve  be  properly  elevated  to  suit  the  velocity.  We  would 
however  remark,  if  the  curve  be  very  short  and  the  velocity  rapid, 
the  friction  of  the  flanges  of  the  wheels  upon  the  outside  rails  of 
the  curve  will  become  so  great  as  to  occasionally  raise  the  face  or 
bearing  of  the  wheel  from  the  rail,  which  falls  again  suddenly  upon 
the  track,  producing  a  perpendicular  jar  or  concussion,  which  may 
sometimes  occasion  lateral  motion  ;  but,  if  the  curves  are  in  good 
order,  and  the  rails  properly  curved,  this  motion  will  be  seldom 
felt,  unless,  I  repeat,  the  radius  be  very  short,  and  the  velocity 
great.  In  this  connection  I  would  add,  it  is  believed  to  be  some 
what  dangerous  to  run  a  train  over  a  curve  in  proper  repair  or  con 
dition  with  a  velocity  so  great  as  to  occasion  this  phenomenon.  We 
have,  however,  felt  this  motion  when  riding  on  curves  not  in  proper 


LATERAL      SHOCKS      ON      A      CURVE.  45 

repair,  or  where  the  rails  have  not  been  curved  in  a  uniform,  regu 
lar  manner ;  whereas,  if  the  track  had  been  in  good  condition, 
nothing  of  the  kind  would  have  been  felt. 

Neither  can  we,  from  any  investigation  we  have  been  enabled  to 
make,  discover  any  reason  for  a  lateral  shock  to  a  car  when  running 
off  of  a  direct  on  to  a  reverse  curve  (if  these  are  the  proper  terms 
of  expression)  provided  the  condition  of  the  track  is  good,  and  the 
wheel  flanges  and  the  rails  are  the  same  distance  apart.  But  the 
question  may  be  asked :  Why  in  practice  lateral  shocks  are  so  fre 
quently  felt,  when  running  from  a  straight  line  on  to  a  curve ;  and 
likewise  when  running  from  a  direct  on  to  a  reverse  curve  ?  We 
answer,  that  the  practice  is  almost  universal  to  lay  the  rails  of  a 
track  from  one  half  to  three  fourths  of  an  inch  further  apart  than 
the  flanges  of  the  car  wheels.  Now  if  we  conceive  the  flanges  of 
the  car  wheels  to  be  in  contact  with  the  line  of  rail  which  forms  the 
inside  line  of  the  curve,  when  the  car  is  about  to  enter  upon  that 
curve,  and  the  track  three  fourths  of  an  inch  wider  than  the 
flanges,  it  will  be  obvious  that  the  motion  of  the  car,  if  it  meets 
with  no  extraordinary  obstruction,  will  continue  straight,  or  in  a 
direct  line,  after  it  enters  the  curve,  until  the  flange  of  the  wheel 
meets  with  the  outside  rail  of  the  curve ;  the  distance  which  the 
car  will  have  then  advanced  into  the  curve,  before  the  phenomenon 
of  contact  takes  place,  will  be  so  great  that  the  curve  will  have 
obtained  a  considerable  degree  of  deflection,  which  of  course  pro 
duces  a  shock,  and  the  shock  will  be  somewhat  proportioned  to  the 
difference  between  the  width  of  the  flanges  of  the  wheels  and  the 
width  of  the  track,  the  velocity  of  the  cars,  and  the  length  of  the 
radius  of  the  curve.  The  following  diagram  (Fig.  7)  indicates  the 
practice  in  cases  of  reversed  curves. 


[FiG.    7.] 


LATERAL      SHOCKS      ON      A      CURVE.  47 

(24)  Let  us  now  examine  the  condition  of  a  car  running  off  of 
a  direct  curve  on  to  a  reverse.  The  centrifugal  force,  as  well  as 
the  position  of  the  wheel  axles,  direct  a  car  when  running  on  a 
curve,  to  the  outside  rail,  which  must,  of  course,  become  its  guide ; 
and  when  the  car  arrives  at  the  reversing  point,  or  rather  one 
fourth  of  its  length  between  its  wheels  beyond  that  point,  it  will 
continue  its  direction  in  a  straight  line  until  its  wheel  flanges  meet 
with  the  outside  rail  of  the  reverse  curve,  and  if  the  difference  of 
width  between  the  flanges  of  its  wheels  and  the  rails  be  consider 
able,  the  car  will  have  advanced  so  far  into  the  curve  before  the 
phenomena  of  contact  take  place,  as  to  admit  the  rail  taking  a 
considerable  degree  of  deflection  ;  and,  as  before  stated  respecting 
simple  curves,  the  meeting  of  the  wheel  flanges  with  the  deflected 
rail  causes  a  lateral  shock  which  is  sensibly  felt,  and  is  proportioned 
to  the  velocity  of  the  cars,  the  difference  in  width  between  the 
flanges  of  the  wheels  and  the  rails,  and  the  length  of  the  radius  of 
the  curve. 

The  motion  of  the  cars  from  the  direct  to  the  reverse  curve  will 
always  be  of  the  same  character  under  similar  circumstances  ;  the 
cars  being  constantly  influenced  by  the  position  of  their  axles,  and 
the  centrifugal  force  of  motion.  But  it  is  not  so  with  the  cars  upon 
a  straight  track ;  there  is  nothing  to  uniformly  guide  them  to  the 
side  of  the  track  which  forms  the  inner  rail  of  the  curve  when  run 
ning  off  of  a  straight  line  on  to  a  curve ;  and,  if  the  flanges  of  the 
wheels  are  in  contact  with  the  side  of  the  track  forming  the  outer 
rail  of  the  curve,  the  car  will  enter  upon  the  curve  without  lateral 
shock. 

If  our  reasoning  has  been  just,  it  would  appear  that  if  the  tracks 


48  USEFUL      FORMULA. 

were  laid  down  to  correspond  with  the  flanges  of  the  wheels,  a  car 
will  meet  with  no  greater  lateral  shock  when  running  through 
reversing  points  of  a  reverse  curve,  than  when  running  off  of  a 
straight  track  on  to  a  simple  curve ;  hence,  under  restricted  circum 
stances,  where  reversed  curves  are  required,  of  short  radius,  it 
hecomes  an  ohject  of  importance  for  the  curves  to  occupy  the  whole 
line,  that  is,  there  should  he  no  straight  track  between  them ;  as  a 
straight  line  of  any  considerable  length  will  tend  to  diminish  the 
length  of  the  radius. 

(25)  Let  us  now  endeavor  to  explain  this  matter  by  diagrams. 
We  will  suppose  a  car  to  be  running  on  a  straight  track  from  A 
towards  F,  with  the  flanges  of  the  wheels  in  contact  with  the  rail 
A  B,  the  flanges  of  the  wheels  on  the  opposite  side  will  not  be  in 
contact  with  the  bar  a  6,  but  will  describe  the  dotted  line  parallel 
with  it ;  the  car  arrives  at  the  tangent  point  B  b,  its  natural 
motion  will  be  from  b  B  to  c  C,  in  the  direction  of  the  dotted  lines, 
where  the  flange  of  the  wheel  running  upon  the  rail  a  f  meets  it  at 
the  point  c,  some  distance  advanced  upon  the  curve,  the  rail  at  this 
point  making  an  angle  with  the  direction  of  the  car,  causes  a  shock 
and  a  sudden  lateral  motion  ;  the  car  then  proceeds  onwards,  with 
the  flanges  in  contact  with  the  rail  a/,  until  it  arrives  at  the  point 
D  d;  from  B  to  D  the  flanges  have  not  been  in  contact  with  the 
rail  A  F,  but  have  described  the  dotted  line  parallel  with  it ;  from 
D  d  its  motion  is  onward  in  the  direction  of  the  tangent  until  it 
arrives  at  E  c,  when  the  flange  meeting  with  the  rail  at  E,  which  it 
will  be  seen  forms  an  angle  with  the  direction  of  the  car,  causes  the 
lateral  shock  felt  at  this  point ;  the  car  then  moves  onwards,  with 
the  flanges  in  contact  with  the  rail  A  F,  until  it  passes  through  or 
over  the  curve,  and  no  further  shock  is  felt.  In  the  mean  time  the 


LATERAL   SHOCKS   ON   A   CURVE.         49 

flanges  upon  the  opposite  side  describe  the  dotted  line  parallel  with 
ef.  If,  therefore,  our  reasoning  be  correct,  it  will  be  obvious  that 
if  we  would  have  cars  run  smoothly  over  a  railroad,  the  track 
should,  near  the  tangent  points  of  curves,  be  laid  down  to  corre 
spond  with  the  width  between  the  flanges  of  the  wheels  ;  and  we 
add  that  the  same  thing  should  be  observed  near  the  turn-out 
frogs,  as  it  is  important  that  the  scores  in  the  frogs  through  which 
the  flanges  of  the  wheels  traverse,  should  be  just  sufficient  to  per 
mit  them  to  pass.  To  render  this  practice  complete  calls  for  a 
greater  degree  of  care  in  the  adjustment  of  wheels  upon  their 
axles,  than  is  at  present  practised  in  many  constructing  and  repair 
shops ;  but,  in  the  present  careless  condition  of  adjustment,  the 
management  of  tracks  can  be  much  improved.  Near  frogs,  and  the 
commencement  of  curves,  the  rails  of  the  track  should  be  no  wider 
than  the  widest  wheels  of  the  train.  A  further  improvement, 
adapted  to  the  passage  through  the  score  in  the  frog  smoothly,  is 
to  have  the  width  between  the  back  side  of  the  wheel  flanges  as 
near  alike  as  they  well  can  be,  which  will  much  improve  the  benefits 
of  the  guard  rails. 

In  this  connection  it  may  not  be  amiss  to  compute  the  deflections 
of  the  rail  at  the  point  where  it  is  met  by  the  flanges  of  the  wheels 
when  running  into  the  curve ;  assuming  the  rails  of  the  track  from 
O'Ol  of  a  foot  wider  than  the  flanges,  up  to  G'06  of  a  foot;  and 
supposing  the  car  to  be  tracing  the  inner  rail  on  its  approach  to  the 
tangent  point,  and  the  curve  to  be  of  a  radius  of  1000  feet. 

The  following  investigation,   and  its  examples  of  computation, 
by  referring  to  Fig.  6,  may  enable  the  student  to  master  the  whole 
merits  of  this  subject. 
H 


50  USEFUL      FORMULAE. 

INVESTIGATION     OF     FORMULA,    AND     EXAMPLES    OF     COMPUTATION. 

Let  r  =  radius  of  the  curve ;  Ji  the  width  between  the  rails  of 

the  track  ; 
Aii   Ao,  As?  etc.,  the  differences  of  distance  between  the  tracks 

and  the  flanges  of  the  wheels  ; 
CD  c-i,  c3?  etc.,  the  corresponding  angles  of  deflection. 

Then,  putting    r  =  1000  feet ;  A  =  4-7  feet;  Ai>   A2,  A3> 

etc.  =  0-01,  0-02,  0-03  feet,  etc.,  we  have 
r  +  J  A  :  B  ::  r  +  i  A  —  An  etc-  :  cos-  '  =  "  +  ^-^'etc- 

Thus,      r  -f  £  A  =  1002-35  co.  ar.     log.  =  6-9989806 

r  -|-  £  7i  —  Ai  =  1002-34  " log.  =  3-0010150 


ci  =  0°  15'  26"  cos.  =  9-9999956 

r  -J-  £  A  =  1002-35  co.  ar.     log.  =  6-9989806 

r  -f  £  /t  —  A2  =  1002-33  log.  =  3-0010107 


c2  =  0°  2L/  44"  cos.  =  9-9999913 

r  -f-  £  A  =  1002-35  co.  ar.     log.  =  6-9989806 

r  +  £  A  —  A3  =  1002-32  log.  =  3 .0010064 


co  =  0°  26'  34"  cos.  =  9-9999870 

r  -\-  ^  h  =  1002-35  co.  ar.     log.  =  6-9989806 

r  +  £  h  —  A4  =  1002-31  log.  =  3-0010020 


c4  =  0°  30'  45"  cos.  =  9-9999826 

r  -f-  -L  A  =  1002-35  co.  ar.     log.  =  6-9989806 

r  +  M  —  As  =  1002-30  log.  =  3-0009977 


cs  =  0°  34'  20"   '          cos.  =  9-9999783 

r+$h  =  1002-35  co.  ar.     log.  =  6-9989806 

r  +  £  h  —  Ae  =  1002-29  log.  =  3-0009934 

ce  =  0°  37'  35"  COS.  =  9-9999740 


ELEMENTS      OF      REVERSED      CURVES.  51 

The  foregoing  computed  deflections  of  the  track,  at  the  point  met 
by  the  wheel  flanges,  under  the  assumed  circumstances,  show  the 
necessity  of  narrowing  the  guage  near  the  commencement  of 
curves,  and  near  the  reversing  points  in  reverse  curves.  There 
can,  however,  be  no  doubt  but  the  cars  will  run  steadier  and  safer 
over  a  narrow  track,  just  suiting  the  wheel  flanges,  than  over  one 
of  greater  width.  The  only  argument  in  favor  of  the  guage  of  a 
track  being  wider  than  the  flanges  of  the  wheels  is,  that  a  greater 
surface  of  the  wheel  is  exposed  to  wear  upon  the  rails  by  the  zig 
zag  course  which  the  wide  guage  allows  the  cars  to  take,  than 
would  be  if  that  motion  was  prevented. 

(20)  The  next  form  of  reverse  curves  which  we  shall  consider, 
is  that  which  shall  unite  two  lines  having  different  bearings,  which 
of  course  would  intersect  each  other  were  they  continued ;  but,  on 
account  of  avoiding  some  obstacles,  or  the  desire  of  a  near  approach 
to  some  particular  locality,  it  becomes  necessary  to  connect  these 
lines  by  reverse  curves ;  it  is,  therefore,  a  matter  of  great  import 
ance  to  lay  down  these  curves  in  the  best  form  possible,  particularly 
if  they  require  short  radii,  which  will  be  best  accomplished,  if 
there  be  no  obstacle  in  the  way,  by  making  the  curves  of  equal 
curvature,  and  occupying  the  whole  distance  between  the  tangent 
points ;  (these  tangent  points  are  supposed  to  be  fixed  by  the  con 
tour  of  the  surface,  or  some  other  consideration  or  governing  prin 
ciple,  which  cannot  well  be  avoided.) 

(27)  To  proceed  with  the  investigation  of  the  proper  formula 
for  determining  the  elements  of  these  curves,  we  would  first 
remark,  that  this  problem  requires  the  angles  A  T'  T  and  T'  T  B, 
and  also  the  length  of  the  line  T  T  to  be  measured.  (See  Fig.  8.) 


ELEMENTS   OF   REVERSED   CURVES.        53 

Then,  by  putting  T  X'  =  c  ,  T  X  =  c,  C  X  =  a,  C'  X  =  b,  and 
T  T  as  measured  =  m,  and  the  radii  C  T  and  C'  T'  in  proportion 
=  unity  ;  (these  radii  by  the  problem  being  equal,)  and  the  radius 
in  measure  =  r. 

Then,  by  the  problem  we  have  the  line  C'  C  =  twice  the  length 
of  radius,  and  it  will  be  apparent  by  a  glance  at  the  diagram  that 
the  angles  X  and  X'  must  be  equal. 

Now,  commencing  with  the  radius  =  unity,  we  have 
Sin.  X  :  1  ::  sin.  (T  —  00°)  :  a=   •i°-,^^90°) 
Sin.  X'  :  i  ::  sin.  (T'-  90°)  :  I  =  sl°-^-,90°) 


Substituting  for  a  -\-  b  their  value,  viz.,  twice  radius,  and  as  we 
have  taken  radius  =  unity,  we  have  this  equation, 

Sin.  (T  —  90  Q)       .       sin.  (T  —  9QQ)    _  _    9 
Si^X  BinTX 

Multiplying  by  sin.  X,  we  have 

Sin.  (T  —  90°)  +  sin.  (T'—  90°)  =  2  sin.  X  ;  hence 

Sin.   X  =    sin-(T-9Q°)  +  sin.(T'-9QQ)  /g-jx 

Having  found  the  angle  X,  we  next  have 

180°—  (T  —  90°)  —  X  =  C  ;  and  180°—  (T'—  90°)  —  X  =  C' 

Then,  Sin.  X  :    1  :  :  sin.  C    :  c  =  -^  (22) 

Sin.  X  :   1  ::  sin.  C'  :  c'  =  -^  (23) 

And,  c  +  c    :  m  :  :      1      :  r  =  -^-  (24) 

Having  thus  obtained  the  radius,  and  the  angles  required,  the 
remaining  elements  necessary  for  making  or  laying  out  the  curve 
may,  of  course,  be  computed  by  such  of  the  foregoing  formulae  as 
the  condition  of  the  locality  requires. 

(28)  NOTE.  Because  of  the  scarcity  of  extensive  tables  of  nat 
ural  sines,  and  for  the  purpose  of  showing  how  readily  they  can  be 


54  USEFUL      FORMULA. 

obtained  from  logarithmic  sines,  we  have,  in  the  specimens  of  com 
putation  given  below,  obtained  the  natural  sine  from  the  loga 
rithmic  sine,  and  after  having  found  the  natural  sine  of  X,  we  have 
deduced  its  logarithm,  and  then  ascertained  the  corresponding 
angle  from  the  tables  of  logarithmic  sines. 

We  have  deemed  it  proper  to  give  the  above  hints,  for  the  inform 
ation  of  such  young  engineers  as  may  not  be  familiar  with  the 
principles  of  trigonometrical  tables. 

To  proceed  with  the  examples  of  computation,  we  have 

T    —  90°  =  87°  52'  58"     log.  sin.  =  9-9997034    nat.  sin.  =      0-9993172 
T'  _  goo  =  740  06'  17"    loff.  sin.  =  9-9830685     nat.  sin.  =      0-9617640 


Nat.  sin.  (T  —  90°)  -f-  nat.  sin.  (T'  —  90°)  =  2)1 -9610812 

X  =  78°  40'  42"     log.  sin.  =  9  9944656     nat.  sin.  =      0-9805406 

We  have  found  X  =  78°  40'  42"  Again,  X  =  78°  40'  42" 

and  (T  —  90°)        =  87°  52'  58"  and  (T'—  90°)  =  74°  06'  17" 

Wherefore  C  must  =  13°  26'  20"  /.  C  must  be      =  27°  13'  01" 


Proof  180°  00'  00"  Proof  180°  00'  00" 

By  (22)  we  now  have  X  =  78°  40'  42"  co.  ar.  sin.  =  0-0085344 

C  =  13°  26'  20"  sin.  =  9-3662513 


c    =     0-2370204  log.  =  9-3747857 

By  (23)  =  again          X  =  78°  40'  42"  co.  ar.   sin.  =  0-0085344 

C'  =  27°  13'  01"  sin.  =  9-6602591 


c'    =0-4664375  log.  =  9-6687935 

By  (24)  c  -|-  c'  =  0-7034579  co.  ar.       log.  =  0-1527619 

ra  =  225-35   feet  log.  =  2-3528576 


r  =320-346    "  log.  =  2-5056195 


PROPORTIONAL      RADII      TO      REVERSED      CURVES.     55 

Having  ascertained  the  angles  and  the  radius,  we  do  not  think  it 
necessary  to  extend  further  the  examples  of  calculation  for  this 
particular  case. 

(29)  We  sometimes  have  another  form  of  the  reverse  curve, 
which  we  will  endeavor  to  investigate.  It  sometimes  happens  that 
in  a  condition  of  the  tangent  lines  not  very  unlike  the  last  de 
scribed,  we  have  some  particular  points  which  we  are  desirous 
should  govern  the  location  of  the  track.  This  state  of  things 
necessarily  fixes  the  length  of  one  of  the  radii,  and  it  is  our  object 
in  the  investigation,  to  deduce  a  formula  for  ascertaining  the  length 
of  the  other,  with  the  centre  angles  which  measure  the  arcs,  etc. ; 
it  being  apparent  that  fixing  the  length  of  the  radius  of  one  of  the 
curves,  governs  the  radius  of  the  other. 

To  proceed  with  the  investigation.  Let  A  T  and  T  B  represent 
the  tangent  lines,  (see  figure  9  ;)  T'  and  T  the  measured  angles, 
viz.,  A  T'  T  and  B  T  T';  and  T'  T  the  line  measured,  which  we 
represent  by  m;  then, 

Putting  a  for  the  line  C'  E  ==  R  T  or  rather  =  D  T  +  K  D 

b      "         "     E  T  =  C'  K 
"         r  =  the  given  radius 
"         x  =  the  radius  required 

cf  =  the  line  R  D 
we  have  a2  +  (b  +  x)2  =  (r~+x)* 

Expanding  the  equation,      a2  +  b2  +  2  bx  +  x*  =  r2  +  2  rx  +  x2 
Subtracting  x2  leaves  a2  -f-  b2  +  2  bx  =  r  -\-  1  rx 

Trans,  and  changing  signs  a2  +  b2  —  r  =  "1  r.r  —  2  bx 

Dividing  by  2  J*!±^zli.  =  (r  —  b)  x 

and  by  (r  -  6)  "i  - «  (25) 


[Fia.     9.] 


COMPUTATION      OF      ELEMENTS. 


57 


Having  deduced  the  formula,  we  proceed  to  give  a  specimen  of 
calculation.      We  will  suppose  m  =  630  feet 

r    =  555     " 
T'  =  169°  30' 
T    =157°  00' 

To  ascertain  a  we  have  first  to  find  T  D ;  in  the  triangle  T  T'  D 
we  have  the 


Angle  T  =  (180°  —  157°)  =  23°  00'  00" 
T'  =  (169-30  —  900)  —  790  so/  00" 
D  =  Supplement  =  77°  SO7  00" 


D       =  77°  30'  co.  ar.    sin.  =  0-0104185 
m       =  630  log.  =  2-7993405 


T'       =790  30' 


sin.   =  9-9926661 


Proof                           =  180°  00/  00" 

TD   =  634-4905 

feet 

log.  =  2-8024251 

m 

log. 

Scno'TKnA 

Sin.  D 

•  ouy  /oyu 

T 

23°    00' 

sin. 

=   9-5918780 

T'D 

252-1373 

=   2-4016370 

T                               == 

555-0000 

In  triangle  CDR,  r  —  T'D=CD  = 

302-8627 

log. 

=   2-4812458 

D 

77-30 

sin. 

=   9-9895815 

b                     = 

295-6837 

log. 

=   2-4708273 

D 

77°   30' 

cot. 

=   9-3457552 

cf                    = 

65-5515 

log. 

=    1-8165825 

From  above                        T  D               = 

634-4905 

(T  D   -f  c')   =                a 

700-0420 

log- 

=   2-8451241 

2 

a2                     = 

490058-8   feet 

log. 

=    5-6902482 

b                       = 

295-6837 

log. 

=  2-4708273 

2 

6*                              = 

87428-82  feet 

log. 

=  4-9416546 

r                       = 

555-00 

log. 

=  2-7442930 

2 

=   308025-00   feet          log.  =   5-4885860 


[FiG.     10.] 


VARIETIES      OF      REVERSED      CURVES.  59 

az   _}_   J2   _  r2      _     269462-62  log.  =   5-4304985 

2    (r  >—  b)  =      5 18- 6326  co.  ar.       log.  =    7-2851402 


x  =     519-5636   feet         log.  =   2-7156387 

b  =     295-6837 


x  -|-  b=  E  C       =      8 15- 24 73*  co.  ar.    log.  =   7-0887106 
a  =     700-0420  log.  =   2-8451241 


C  =  40°   39'    08" -09        tan.  =   9-9338347 

T/  oo  T  =    12°    30'    00". 00 


C'  =    28°    09'    08" -10 

Having  found  the  centre  angle  C,  we  can  readily,  by  several 
methods,  ascertain  the  value  of  the  other  centre  angle,  C;  we  how 
ever  shall  only  give  the  following  method,  viz. : 

When  the  angle  D  in  the  triangle  T'  D  T,  is  greater  than  a  right 
angle,  the  difference  between  the  angles  A.T'  T  and  B  T  T  must 
he  added  to  C,  and  the  sum  will  be  equal  to  C';  and  when  D  is 
smaller  than  a  right  angle,  it  must  be  subtracted. 

The  remainder  of  the  elements,  which  may  be  needed  to  facilitate 
the  operations  of  location,  may  without  difficulty  be  found,  by  such 
of  the  foregoing  formulae  as  shall  be  found  applicable. 

(30)  Another  form  of  the  reverse  curve  is  wherein  we  may 
have  one  tangent  point  fixed,  and  one  centre  angle  given,  the  tan 
gent  lines  being  located  in  position  and  direction,  but  one  of  them 
may  be  shortened  or  lengthened  to  adapt  it  to  the  unknown  or 
required  angles.  This  case  occurs  when  the  point  where  the  curves 
reverse  becomes  a  governing  point  in  the  location,  as  at  the  point  G 
in  the  figure. 

*  Having  in  the  triangle  C'  C  E   found  the  sides,  C'  E   =  a,  and  C   E  =  x  +  6,  we  have  to 

find  the  angle  C  ;          "    --    =  tan.  C. 
x  -f-  o 


60  USEFUL     FORMULA. 

INVESTIGATION     OF     FORMULAE. 

In  the  diagram  the  tangent  lines  are  represented  by  A  T  and 
B  T ;  and  the  angles  A  T'  T  and  B  T  T'  by  Tm  and  Tm  ;  the  line 
measured,  viz.,  T  T  by  m. 

Then,  putting  t'  for  the  line                    D  T 

And,                  t  "  "  "                     D  T' 

r  "  "  given  radius     C'  T' 

r  "  "  radius  sought  G  C 

We  have  in  the  solution  of  this  problem  the  following  triangles, 
viz.,  T'  D  T,  which  for  convenience  we  denominate  No.  1 ;  C'  D  E, 
No.  2  ;  E  F  G,  No.  3 ;  F  G  C,  No.  4 ;  which  require  to  be  succes 
sively  solved. 

Commencing  with  triangle  No.  1,  we  have 
The  angle  at  T  =  180°  —  Tm 
«     «  T'  =  Tm    •  -  90° 
"     "  D  =  the  supplement  of  the  above. 

Then,  by  analogy,  Sin.  D  :  m  : :   sin.  T   :  t  =  -^^D—  (2G) 

And,  Sin.  D  :  m  : :  sin.  T'  :  t'  =  -^j-  (27) 

In  the  solution  of  triangle  No.  2,  we  have  the  side  C'  D  =  r  —  t, 
and  the  angle  D  =  the  supplement  of  D  in  triangle  No.  1 ;  the 
angle  C'  being  given,  the  angle  E  =  the  supplement  of  C'  -f~  D- 
Denoting  the  side  C  E  by  d,  and  the  side  D  E  by  c',  we  have 

Sin.  E  :  r  —  t  ::  sin.  D  :   d  (28) 

and  Sin.  E  :  r  —  t  ::   sin.  C'  :    c  (29) 

In  triangle  No.  3  we  have  r  —  d  =  E  G ;  the  angle  E  the  same 
as  E  in  No.  2  ;  the  angle  G  a  right  angle ;  and  of  course  the  angle 
F  becomes  the  complement  of  E.  Denoting  the  side  G  F  by  c,  we 


COMPUTING      ELEMENTS.  61 

have         Cos.  E  :  r  —  d  ::   sin.  E  :  e  =  tan.  E  (r  —  d) 
Then,  in  the  quadrilateral  F  G  C  T,  we  have  the  angle  at  F  = 
(180° — F),  F  being  the  same  as  in  No.  3;  bisecting  F  as  thus 
found  by  a  line  from  F  to  C,  we  have  in  triangle  No.  4, 

Cos.  \  F  :  e  : :   sin.  i  F  :  r  =  tan.   |  F  .  e  = 

tan.  I  F  .  tan.  E  .  (r  —  d)  (30) 

The  last  expression  being  equal  to  the  side  G  C. 

In  practice  it  will  be  convenient  to  ascertain  the  distance  of  the 
tangent  point  from  the  point  T,  and  for  that  purpose  we  have  in  the 
triangle  C  T  E,  which  we  call  No.  5, 

Sin.  E  :  /  : :  cos.  E  :  a  =  cot.  E  .  /  (31) 

where  c\  represents  the  line  E  T. 

Then,  will  (c  +  ci)  ^  t'  =  the  distance  of  the  tangent  from  the 
point  T,  and  the  direction  of  course  will  be  known  from  the  relative 
magnitude  of  the  numbers  represented  by  (ci  +  c)  and  t' ,  viz.,  if  t' 
represent  the  larger  number,  the  tangent  point  will  be  in  the  direc 
tion  of  D ;  if  the  smaller,  in  the  direction  of  B. 

EXAMPLE      OF      COMPUTATION. 

Suppose  m  =  030  feet ;  r  =  555  feet ;  Tm  =  109°  30';  Tm  = 
157°  00';  C'=  27°  49'  27" -35. 

Then,        T  =  (180°  —  Tm  )  =  180°  -  157°  =  23°  00' 

T'  =  (Tm  -  -  90°)  =  169°  30'  -       90°  =  79°  30' 
D  =  supplement  77°  30' 

By  formula     (26)    D          =  77°  30'  co.  ar.     sin.   =  0-0104185 

m  =  630  feet  log.  =  2-7993405 

T          =  23°  sin.   =  9-5918780 


t  =  252-1373  log.  =  2-4016370 


62 


USEFUL     FORMULAE. 


(27)    D 

M 

T/ 
t> 


=  77°  30> 

=  630  feet 

=  79°  30' 

=  634-4905  feet 


co.  ar.  sin.  =0-0104185 
log.  =  2  7993405 
sin.  =  9-9926661 


lo£T.  =  2-8024251 


(28)    E 

555  —  252-1373  =  r  —  t  =  302-8627  feet 
D  =102°  30' 


=  49°  40'  32"- 65  co.  ar.  sin.  =  0-1178203 
log.  =  2-4812458 
sin.  =  9-9895815 


d 

=  387- 

836 

feet 

log. 

=  2- 

5886476 

(29) 

E 

=  49° 

40' 

32".  65  co.  ar. 

sin. 

=  0- 

1178203 

r  —  t 

=  302- 

8627  feet 

log. 

=  2-4812458 

C' 

=  27° 

49/ 

27"-35 

sin. 

=  9- 

6690948 

c' 

=  185- 

4219  feet 

log. 

=  2- 

2681609 

(30) 

»» 

=  69° 

50' 

16".  32 

tan. 

=  o. 

4351232 

E 

=  49° 

40' 

32".  65 

tan. 

=  0- 

0711997 

r  —  <f 

=  167- 

164 

feet 

log. 

=  2- 

2231428 

r' 

=  536- 

3715  feet 

log. 

=  2- 

7294657 

(31) 

E 

=  49° 

40' 

32".  65 

cot. 

=  9- 

9288003 

r; 

= 

log. 

=  2- 

7294657 

c' 

=  455 

•  267  feet 

log. 

=  2- 

6582660 

Cl 

=  185 

-422 

c'  +ci   =  640-689 

t  =  634-490 

(ci  +c')  c^t  = 


6-  199  feet  = 
the  distance  the  tangent  point  must  be  fixed  from  T  towards  B. 

Again,  suppose  m  =  630  feet;  r  =  555  feet;  Tm  =  169°  30'; 
Tm=  157°;  C'  =  28°  09'  08"  -1. 

Then,  T  =  160°  -  157°  =  23°  00' 

T  =  169°  30'  -       90°  =  79°  30' 
D  =  supplement  =77°  30' 

180°  00' 


TESTING      COMPUTATIONS.  63 


(26) 

D 

=  77° 

30' 

co.  ar.  sin. 

=  0- 

0104185 

in 

=  630 

feet 

log. 

=  2- 

7993405 

T 

=  23° 

00' 

sin. 

=  9- 

5918780 

t 

=  252 

•  1373  feet 

log. 

—  2- 

4016370 

(27) 

D 

=  77° 

30' 

co.  ar.  sin. 

=  0- 

0104185 

m 

=  630  feet 

log. 

—  2. 

7993405 

T/ 

=  79° 

30' 

sin. 

=  9- 

9926661 

t' 

=  634- 

4905  feet 

log. 

=  2- 

8024251 

(28)    E          =  49°  20'  51". 9    co.  ar.     sin.  =  0-1199428 
r  —  t  =  302-8627  feet  log.  =  2-4812458 

D          =  102°  30'  sin.   =  9-9895815 


d           =  389-7356  feet  log.  =  2-5907701 

(29)    E          =  49°  29'  51". 9  co.  ar.      sin.   =  0-1199428 

r  —  t  =  302-8627  feet  log.  =  2-4812458 

C'         =  28°  09'  08"- 1  sin.  =  9-6737728 


c'  =  188- 3482  feet  log.  =  2-2749614 

(30)  |F      =     69°  40'  25"- 95  tan.  =  0-4312942 
E          =     49°  20'  51"- 90  tan.  =  0-0661653 

r  =  555  ;  d  =  389-7356,     r  —  d  =  165°  26'  44"  log.  =  2-2181792 

r  =  519-5636  feet  log.  =  2-7156387 

(31)  E  =     49°  20'  51". 9  cot.  =  9-9338347 


c  =  446-1423  feet  log.  =  2-6494734 

c'          =  188-3482 


<*  +c=  634-4905 
t  =  634-4905 

In  this  example  it  appears  that  the  tangent  point  sought  is  at  T. 

NOTE.  The  reader  will  perceive  that  this  example  is  taken  from 
the  results  of  the  example  next  hut  one  preceding,  and  is  intended 
as  a  test  to  hoth. 


[FIG.      11.] 


INSTRUCTIVE     EXAMPLE     OF     REVERSED     CURVES.        65 

(31)  Another  form  of  reversed  curve  (if  the  expression  be  a 
proper  one)  for  uniting  tracks  having  different,  or  like  bearings,  is 
where  you  have  the  relative  position  of  the  tangent  points  from 
whence  the  curves  commence  in  the  given  tracks,  with  the  bearings 
of  said  tracks,  or  (which  is  the  same  thing)  the  tangent  lines,  and 
the  radii  of  the  curves  given  or  required  by  the  contour  of  the  sur 
face,  or  other  considerations  which  may  govern  the  location. 

In  order  to  give  this  problem  a  practical  character,  we  copy  from 
a  case  which  actually  occurred  in  the  practice  of  the  writer.  We 
shall  not  give  all  the  preliminary  surveying  which  was  deemed 
necessary  to  guide  us  in  the  location,  (which  had  been  in  amount 
considerable,)  but  will  only  state  that  many  lines  were  run  and 
measured  in  various  directions,  to  such  points  as  we  were  desirous 
of  knowing  the  relative  situations  of,  and  the  traverses  were  worked 
up ;  or,  in  other  words,  the  relative  situations  of  these  points  were 
computed  in  northings  and  southings,  eastings  and  westings.  We 
copy  from  those  tables  such  data  as  we  shall  find  necessary  to 
enable  us  to  explain  and  solve  the  problem,  and  render  our  com 
putations  intelligible. 

The  position  of  first  tangent  point,  174  -306  feet  northing,  159*617 

feet  easting. 
The   bearing   of    tangent   line   from   first   tangent    point,   N.    E. 

85°  44'  35". 
The  bearing  of  radius  from  first  tangent  point,  S.  E.  4°  15'  25", 

and  its  length  =  503 -118  feet,  log.  =  2. 7016699 

The   position   of   the  centre   of  the  curve  formed  by  the   above 

radius  =  327424  feet  southing,  196-963  feet  easting. 
The  position  of  second  tangent  point,  64-735  feet  northing,  509-235 

feet  westing. 


66  USEFUL      FORMULA. 

The  bearing  of   tangent  line  from  second  tangent  point,  X.  W. 

61°  11'  53". 
The  bearing  of  radius  from  second  tangent  point,  X.  E.  28°  48'  07" 

and  its  length,  306  •  705  feet,  log.  =  2 •  5984678 

The  position  of   the   centre  of   the   curve   formed   by  the   above 

radius  =  412-364  feet  northing;  318-109  feet  westing. 

Representing  the  first  tangent  point  by  T!  ,  and  the  second  tan 
gent  point  by  T2?  and  the  interior  tangent  points  by  T'  and  T', 
(see  Fig.  11,)  the  centre  of  the  curve  with  the  radius  of  503-118 
feet  by  C,  and  the  centre  of  the  curve  with  the  radius  of  396  •  705 
feet  by  C';  then  will  C  T  =  503-118  feet,  and  C  T'  =  396-705 
feet. 

Having  constructed  our  diagram  in  conformity  with  the  data 
given,  we  commence  by  finding  the  distance  C  C'.  We  have  given 

the 

Position  of  C    =  327-424  feet  southing,  and  196-963  feet  easting, 

«        «    c/  =  412-364  feet  northing,  "     318-109  feet  westing, 

Diff.  of  northing,    739-788  feet  515-072  feet  diff.  of  westing. 

Having  obtained  the  difference  of  northings  and  westings  between 
C  and  C",  we  have  this  analogy,  to  find  the  bearing  from  one  to  the 
other,  viz.  Assuming  the  distance  C  C'  =  radius,  and  the  differ 
ence  in  the  northings  as  a  cosine,  and  the  difference  of  westings  as 
a  sine;  then,  representing  these  functions,  viz.,  radius  by  E,  the 
sine  by  «,  the  cosine  by  c,  the  bearing  by  B, 

We  have     c  '.  s  '.'.   R  I  tan.  B  =  -| 

or,  more  practically  to  express  the  same  thing,  we  have,  tan.  B  = 
difference  of  westings,  divided  by  the  difference  of  northings,  and 

Sin.  B  :  westing     : :  R  :  C  C'  =  -^jr 
or,  Cos.  B  :  northin      : :  R  :  C  C'  = 


INSTRUCTIVE     EXAMPLE     OF     REVERSED     CURVES.          67 

Having  thus  found  C  C',  we  compare  it  with  the  sum  of  the 
radii,  viz.,  C  T'  +  C'  T";  and  if  C  T'  +  C'  T"  be  found  greater 
than  C  C',  the  assumed  radii  will  be  too  great  for  the  relative  situ 
ation  of  things ;  but  if  C  T  +  C'  T"  be  found  less  than  C  C'  the 
two  curves  will  not  run  into  each  other,  and  must  be  connected  by 
a  piece  of  straight  line.  The  determination  of  this  straight  line  is 
the  object  of  the  present  investigation.0 

For  an  example  of  computation  we  have  given  the 

Position  of  C    =  327-424  feet  southing,  and  196  963  feet  easting, 

and  of  C'  =  412-364    "    northing,  "     318-109    "    westing, 

Diff.  of  northings,  739-788    "  "     515-072    "    diff.  of  westings. 

Difference  of  westings  =515-072     log.  =  2-7118680 

"  "    northings          =739-788     log.  =  2-8691072 


Bearing  from  C  to  C'  =  B  =  N.  W.  34°  50'  50". 24  tan.  =  9-8427608 

B  =  34°  50'  59" -24  co.  ar.  sin.   =  0-2430665  CO.  ar.    cos.  =  0-0858273 

Westing  =515-072    log.  =  2-2118680     northing  739 - 788     log.  =  2-8691072 


From  C  to  C'  =  901-435     log.  =  2-9549345  Proof,  log.  =  2-9549345 

The  given  radii  are 


From  Ti  =503-118  feet 

"       T2  =  396-705 


C  T'  +  C  T"  =  r  -f-  r  =  899-823 
C  C'  =  901-435 

Difference  1-6 12  feet 

C  C'  being  longer  than  C  T'  -(-  C'  T",  it  is  evident  it  will  require 
several  feet  of  straight  line  to  connect  them. 

*  It  is  obvious  that  if  C  C'  and  (C  T'  +  C'T")  are  found  the  same  length,  the  two  will  run  into 
each  other,  and  form  perfect  reverse  curves. 


68  USEFUL      FORMULA. 

Eepresenting  C  T'  +  C'  T"  by  r  +  r  ,  and  C  C'  by  H,  we  have 

H   :  E         : :  r  +  /  :  sin.  C  or  sin.  C'  =  ^-^ 

Sin.  C    :  r+r  ::  cos.  C  •  DC  =  D'  C'  ==  T'  T"  =  cot.  C  (r 

.-.  r  -f  r'  =  899-823       log.  =  2-9541571 

H     =  901-435  CO.  ar.  log.  =  7-0450655 


C     =  86°  34'  22"    sin.  =  9-9992226 

And   C     =  86°  34'  22"    cot.  =  8-7772380 

r  -|-  r  =  899-823       •••-  log.  =  2-9541571 


T/ T"  =  53-888  feet    log.  =  1-7313951 

Having  thus  found  the  length  of  the  straight  line  connecting  the 
two  curves,  it  becomes  a  matter  of  considerable  interest  to  know  the 
magnitude  of  the  centre  angle  belonging  to  each  curve  respectively. 

We  found  the  bearing  from  C  to  C'  to  be  N.  W.  34°  50'  50" -24  ; 
and  the  angle  C  in  the  triangle  C  C'  D  =  8G°  34'  22",  the  comple 
ment  to  which  will  be  3°  25'  38". 

Then,  the  bearing  from  C  to  C'   =  N.  W.  34°  50'  50'/-24 

It  is  obvious  that  if  we  subtract  the  complement  D'  C  C'  =  3°  25'  38" 


Will  leave  the  bearing  C  T'  and  C'  T" =  N.  W.  31°  25'  12"  -24 

The  bearing  C  Ti  being =  N.  W.    4°  15'  25" 


Gives  the  centre  angle  Ti  C  T'  =  27°  09'  47"-24 

Again,  C'  T"  bearing  (as  above) =  N.  W.  31°  25'  12"-24 

And  the  radius  T2  C  bears =  N.  E.    28°  48'  07" 


Gives  for  the  centre  angle  Ta  C'  T" =  60°  13'  19". 24 

Such  further  elements  as  may  be  deemed  useful  in  the  location 
might  be  readily  obtained  by  such  of  the  preceding  formula  as  may 
be  found  applicable. 


SWITCH-BARS.  69 

(32)  There  will  doubtless  arise  in  practice  a  great  variety  of 
cases,  or  conditions  requiring  reverse  curves,  many  of  them  requir 
ing  formula   entirely  different  from  those  we  have  been  investi 
gating,  while  there  are  many  others  which  will  require  merely  some 
slight  modifications.     But,  to  repeat  what  has  been  more  than  once 
stated,  it  is  not  our  purpose  to  pursue  these  investigations  until  the 
subject  is  exhausted,  but  only  to  present  those  cases  which  we  have 
presumed  would  most  frequently  occur  in  practice.     We,  however, 
have  another  class  of  curves,  the  greater  portion  of  them  reversing 
curves,  viz.,  turnouts  and  side  tracks,  which   may  be  worthy  of 
consideration.     We  will  therefore  proceed  to  the  investigation  of 
formulae  for  obtaining  the  necessary  elements  for  locating  them, 
and  in  the  same  connection  will  endeavor  to  ascertain  the  magnitude 
of  the  angles  the  rails  make  with  each  other  at  the  points  of  crossing, 
or,  in  other  words,  the  dimensions  and  form  of  the  frogs  necessary 
to  be  used  to  best  suit  each  particular  case. 

(33)  Before  we  proceed  with  the  investigations,  I  would  make 
a  few  remarks  upon  the  switch-bar.     The  switching  of  the  bar, 
preparatory  to   turning   a  train   upon  a  side   track,  becomes   an 
important  element  in  our  investigation.     We  have  no  doubt  this 
element  would  be  considered  by  persons  who  have  not  fully  investi 
gated  the  subject,  as  unnecessarily  complicating  our  formulae,  and 
of  course  our  computations. 

The  first  consideration  in  preparing  the  switch,  is  to  ascertain 
the  smallest  amount  of  sliding  motion,  that  will  answer  to  pass  the 
wheels,  and,  at  the  same  time,  give  firmness  and  security  to  the 
ends  of  the  rails.  The  pattern  of  rails  generally  used  in  Massa 
chusetts  requires  a  movement  of  about  five  inches,  and  the  pattern 


70  USEFUL      FOBMULJE. 

for  the  switch  castings  used  to  secure  the  ends  of  the  rails,  and  to 
give  firmness  and  stability  to  the  structure,  are  nearly  uniform  in 
their  dimensions ;  hence,  whether  the  switch  rail  be  long  or  short, 
whether  the  turnout  be  of  large  or  small  radius,  the  switching,  or 
movement  of  the  end  of  the  bars,  remains  the  same. 

It  may  happen,  however,  that  when  the  turnout  curve  is  required 
to  be  exceedingly  severe,  and  we  desire  to  make  the  most  of  the 
room  we  have  at  command,  that  we  determine  by  calculation  the 
length  the  switch-bar  (switching  five  inches,  or  the  amount  required 
by  the  castings)  must  be  to  make  it  exactly  correspond  to  a  portion  of 
the  intended  curve,  and  the  switch  rails  are  accordingly  cut  to  that 
length.  But,  if  there  is  nothing  to  prevent  the  radius  of  the  turn 
out  from  taking  such  length  as  may  be  deemed  most  desirable,  it 
becomes  the  better  policy  to  have  the  switch-bars  as  long  as  the  bars 
with  which  the  track  is  laid,  or  is  being  laid. 

The  longer  the  switch-bar  is,  the  smaller  will  be  the  angle  of 
deflection  occasioned  by  switching ;  and  the  smaller  the  deflecting 
angle,  the  less  the  impediment  to  the  passage  of  the  engine  and 
cars,  and  less  springing  of  the  bar  than  when  the  bar  is  shortened ; 
and  of  course  less  liability  to  accident. 

In  general,  the  deflection  of  the  switch-bar  should  not  be  greater 
than  the  deflection  of  the  curve  for  the  same  length  of  arc.  Cases 
will,  however,  occur,  when  the  deflection  of  the  switch-bars  of  the 
greatest  length  in  use,  will  exceed  the  deflection  of  the  like 
quantity  of  arc.  These  cases  occur  frequently  at  the  connection  of 
branches ;  and,  in  general,  we  may  say  (if  this  discussion  be  cor 
rectly  based)  that  the  switch-bar,  when  switched,  should  in  all  cases 


SWITCH-BAKS.  71 

be  considered  the  tangent  line  from  whence  the  curve  is  to  spring, 
or  commence.  It  may,  however,  be  neglected,  when  the  switching 
exactly  corresponds  with  the  deflection  of  the  same  length  of  curve ; 
but  it  will  not  in  that  case  interfere  with  the  accuracy  of  the  calcu 
lations  to  then  consider  it  as  the  tangent  line.  Cases  may,  however, 
occur,  requiring  the  tangent  lines  to  be  continued  beyond  the  end 
of  the  bar  before  the  curve  commences ;  but  these  cases  will  not 
often  be  met  with. 

(34)  Having  said  thus  much  respecting  switches,  we  commence 
our  investigation  by  considering  the  most  simple  form  of  the  turn 
out,  viz.,  from  a  straight  track,  with  curves  of  equal  radii. 

NOTE.  I  would  here  state,  for  the  information  of  the  young 
engineer,  that  the  side-track  curves,  when  there  is  nothing  to  inter 
fere,  should  be  laid  to  a  radius  of,  say  from  five  to  six  hundred  feet ; 
but  when  the  nature  of  things  demand  it,  they  may  be  laid  to  a 
radius  as  short  as  two  hundred  and  fifty  feet.  If  a  radius  still 
shorter  is  demanded,  it  becomes  necessary  to  lay  extra  rails  upon 
the  inside  of  the  curve,  and  as  near  the  rails  of  the  main  track  as 
they  can  be  well  secured,  to  assist  in  supporting  the  centre  driving 
wheels  of  the  engine,  which  would  otherwise  be  sometimes  unsup 
ported,  and  would  then  cause  the  engine  to  run  off  the  track.  I 
hardly  need  to  remark  that  the  double  rail  will  be  useless  when  the 
engines  have  only  one  pair  of  driving  wheels. 

To  proceed  with  the  investigation.  We  first  ascertain  the  relative 
position  of  the  switch-bar,  or  the  angle  it  makes  with  the  main 
track. 


[Fia.     12.] 


FROG      ANGLES.  73 

Let  S  represent  the  length  of  the  switch-rail,  and  d  the  distance 
it  slides  ;  Sw  the  switch  angle,  or  the  angle  the  switch-rail  makes 
with  the  main  track  when  it  is  switched.  We  then  have 

8:  R  :  :  d  .  sin.  Sw  =  j-  or  tan.  /Sto°  (32) 

Having  thus  obtained  the  switch  angle,  we  will  now  put  r  = 
radius  of  the  turnout  ;  a  =  C  Q  ;  6  =  Q  T;  q  =  C  A  ;  d  =  the 
distance  between  the  track  centres  ;  e  =  AT;</  =  TB. 

We  have  in  the  triangle  A  C  T,  to  find  q  and  e. 

E  :  r  :  :  cos.  Sw  :  a  =  r  .  cos.  Sw  ) 

(33) 
E  :  r  :  :  sin.  Sw  :  e  =  r  .  sin.  Sw  ) 

Then  will  b  —  q  —  &  +  d,  and 

2  r  :  E  ::  (r  +  b)  ;  cos.  C'  =  -^  (34) 

Cos.  C'  :  (r  +  5)  :  :  sin.  C'  :  a  =  tan.  C'  (r  -f  b)       (35) 

We  also  have  a  —  e  =  g;  and  the  angle  C  =  C'  —  $#. 

We  have  now  found  the  principal  elements  necessary  for  locating 
and  marking  the  centre  line  of  the  turnout  ;  whatever  practice 
requires  to  fill  up  the  details  may  readily  be  supplied  from  formulae 
given  in  the  preceding  pages. 

(35)  The  frog  angle  next  claims  our  attention.  Eepresenting 
the  distance  between  the  rails,  or  in  other  words,  the  guage  of  the 
track,  by  h,  we  have  C  F  =  r  +  J  h;  and  C  0  =  r  —  J  h  +  d; 


*  To  be  strictly  exact,  we  make  use  of  the  following  analogy  : 

s  :  R  :  :  j  d  :  sin.  i  sw  =  -~^- 

o 

but  this  formula  is  rather  a  refinement  than  otherwise,  as  either  of  the  two  first  expressions  are 
sufficiently  exact  for  practice,  and  more  convenient. 
L 


74  USEFUL      FORMULA. 

and  the  angle  C  0  F  =  90°  +  $w.     We  now  have,  in  the  triangle 
C  0  F,  r  +  i  h  :  sin.  (90°  +  8w)  : :  r  —  J  h  +  d  :  sin.  F  = 

sin.  (900  -|-  sw)  (r  —  } 


(r  +  i  A) 


(36) 


Then,  drawing  at  F  the  tangent  line  F  M,  which  of  course  must 
he  at  right  angles  to  C  F,  it  will  be  apparent  that  the  frog  angle 
M  F  0  will  he  a  complement  angle  to  F  as  found  above  ;  where 
fore,  we  have  90°  —  F  =  M  F  ©  ;  and  the  angle  at  the  centre  C, 
will  be  equal  to  180°  —  (0  +  F  ;)  or,  which  amounts  to  the  same, 
C  =  M  F  ©  —  Sw;  and  the  chord,  which  we  represent  by  ch,  from 
the  mouth  of  the  switch  upon  the  outside  rail  of  the  turnout  track, 
to  the  point  where  the  frog  angle  should  be  placed  in  the  main 
track,  may  be  ascertained  by  the  following  analogy  : 

Sin.  i  (180°  -  C)  :  r  +  J  A  :  :  sin.  C  :  ch  =        c£±i«-  (37) 


The  chord  just  found  will  be  of  great  convenience  to  the  track 
layers,  as  it  will  show  them  the  proper  place  for  the  frog,  which 
should  be  put  into  the  main  track  when  they  are  laying  it  down. 

Having  thus  obtained  our  formula},  we  now  proceed  with  an 
example  of  computation. 

We  will  assume  r  =  499  '725  feet,  which  gives  a  deflection  of 
1°  2G'  for  a  25  ft.  chord  ;  h  =  4-7  feet  ;  d  =  5  inches;  S=  21 
feet  ;  and  5=11  feet. 

By  formula  (32)  we  have 

d    =  5  inches  ............     log.  =  0-6989700 

S    =  252  inches  co.  ar.      ............     loff.  =  7  •  5985995 


—  1°  08'  12"  tan.  =  8-2975695 


FROG     ANGLES. 

Sw=  1°  08'  12"  cos.  =  9-9999145 

r     =499-725  log.  =  2-6987307 


q     =  499- 6262  feet          log.  =  2-6986452 

Again,           Sw=  1°  08'  12"              sin.  ==  8-2974820 

r     =                                   log.  =  2-6987307 

e     =9-9131                       log.  =  0-9962127 

2  r        =  999  -449  co.  ar.        log.  =  7-0002393 

r  -f-  b  =  988-768                     log.  =  2-9950944 


(33) 


C'         =  8°  23'  02"-7  COS.  =  9-9953337  (34) 

C'  =  8°  23'  02" -7  tan.  =  9-1684391 

r-f6      =  log.  =  2-9950944 

=  145-7248feet  log.  =  2-1635335  (35) 

=   9-9131 


=  135-8117 


the  distance  on  the  main  track  from  the  mouth  of  the  switch  to  a 
point  opposite  T' ,  T  being  off  at  right  angles  from  the  point. 


C'     =    8°  23'  02". 7 

Sw    =    1°  08'  12" 


C'—  Sw  =  C  =    7°  14'  50"- 7 

r  +  i  h             =  502-0746  feet  co.  ar.    log.  =  7-2992318 

90°  +  Sto         =  91°  08'  12"  sin.   =  9-9999145 

,-  _j_  d  —  i  h  =  497-7912  feet  log.  =  2-6970473 

F                         =  82°  25'  31"  sin.   =  9-9961936 

(36) 

9QQ  _  F  =  frog  angle  =  7°  34'  29" 
Sw  =  1°  08'  12" 
Frog  angle  —  Sw  =  C"      =  6°  26'  17";   C"  = 


the  angle  at  C  in  the  triangle  T  C  F. 


76  USEFUL      FORMULA. 

£  (180°—  C)  =  86°  46'  51". 5    co.  ar.    sin.  =  0-0006858 
r  +  £  h  =  502-0746  feet  log.  =  2-7007682 

C"  =  6°  26'  17"  sin.  =  9-0497178 


ch  =  56-386  feet  log.  =  1-7511718  = 

chord  distance  from  mouth  of  switch  to  the  angle  of  the  frog  upon 
the  outside  rail  of  the  turnout. 

Recapitulation  of  the  elements  ohtained,  viz., 

Centre  angles  C'      =     ........................................     8°  23'  02".  7 

«       C       =      ........................................      7°  14'  50"-  7 

Frog  angle  M  F  ©  =      ........................................     7°  34'  29"-0 

Chord  distance  from  the  mouth  of  switch  to  mouth  of  frog,  (outside  rail,)  56-386  feet. 

(36)  We  find  wanting  the  relative  position  of  the  point  where 
the  curves  reverse.     The  formula  will  be 

Sin.  }  (180°  -  C)  :  r  :  :  sin.  C  :  e  = 


wherein  c  =  the  chord  distance  from  the  centre  point  between  the 
mouth  of  the  switch-bars  when  switched,  and  the  point  where  the 
curve  reverses. 

Then,  to  find  the  chord  of  the  reverse  curve  =  c  ',  we  have 

Sin.  i  (180°—  C')  :r  i:  sin.  C'  :  c  =-'^^  (3D) 
representing  by  c  in  the  foregoing  analogy,  the  chord  from  the 
reversing  point  to  the  tangent  point  T. 

EXAMPLE      OF      COMPUTATION. 

^  (180°  —  C)  =  86°  22'  34"-G5  co.  ar.     sin.   =  0-0008692 
r     =  499-  725  feet  log.  =  2-6987,307 

C    =  7°  14'  50"  sin.   =  9-1008914 

c     =63-167  feet  log.  =  1-8004913         (38) 


KELATION      BETWEEN      TURNOUT      AND      FROG.        77 

£  (180°—  C')  =  85°  48'  28" -65  co.  ar.    sin.  =  0-0011635 
r     =  499- 725  feet  log.  =  2-6987307 

C'    =  8°  23'  02"*7  sin.  =  9-1631819 


c'    =72- 958  feet  log.  =  1-8630761          (39) 

(37)  To  lay  down  the  chords  c  and  <?',  we  commence  by  placing 
the  theodolite  at  the  point  in  the  centre  of  the  mouth  of  the  switch, 
when  switched  ;  then,  pointing  the  telescope  in  the  direction  towards 
B,  in  a  range  parallel  with  the  main  track,  lay  off  an  angle  towards 
the  side  of  the  road  to  which  the  switch-bar  switches  =  (90°  —  Sw) 
-f-  5  (180° —  C,)  then  measure  from  the  instrument  the  distance  c 
for  the  reversing  point.    Then,  moving  the  theodolite  to  the  reversing 
point,  and  directing  the  telescope  to  the  point  just  left,  (to  wit,  in 
the  centre  of  the  mouth  of  the  switch,)  lay  off  an  angle  towards 
the  main  track  =  180°  —  \  (180°  —  C)  +  J  (180°—  C',)  and 
measure  the  distance  c  for  the  tangent  point  of  the  turnout.     The 
remainder  of  the  laying  out  may  be  performed  by  deflections,  or 
other  methods,  as  explained  in  the  foregoing  pages. 

We  have  now,  I  think,  obtained  every  element  necessary  for 
locating  and  marking  out  a  turnout  from  a  straight  track,  and  for 
making  a  frog  pattern  to  suit. 

N.  B.  If  the  tangent  points,  the  reversing  point,  and  the  place 
for  the  frog  be  distinctly  and  properly  marked,  and  the  rails 
properly  curved,  a  skilful  tracklayer  would  put  in  a  turnout 
without  further  laying  out. 

(38)  We  now   proceed   to   the   investigation   of   formulae   for 
determining  the  radius  of  a  turnout  from  a  straight  track  suited 
to  a  given  frog. 


[Fio.     13.] 


RELATION      BETWEEN      TURNOUT      AND      FROG.        79 

Let  Sw  represent  the  switch  angle. 

Fr  "  "    frog         " 

C  "  "    centre      " 

A  "  one  of  the  equal  angles  in  isosceles  triangle  AFC. 

h  "  the  distance  between  the  rails. 

d  "  "         "        the  switch  slides. 

r  "  "    radius. 

We  now  have  in  the  triangle  B  C  F,  (See  Fig.  13,)  the  angle 
at  B  =  90°  +  Suo;  the  angle  at  F  =  90°  —  Fr  ;  and  the  angle  at 
C  =  180°  —  (B  +  F.)  Then,  in  the  triangle  A  B  F,  representing 
F  by  F2,  and  B  by  B2,  we  have  the  angle  A  =  i  (180°  —  C  ;)  and 
the  angle  F2  =  (A  —  F,)  F  being  =  (90°  —  Fr)  as  above,  and 
the  angle  BS  =  90°  —  Sw. 

Having  thus  determined  the  angles,  we  have  in  the  triangle 
FAB,  A  B  =  ^£-;  that  is,  sin.  B2  :  h  —  d  I :  B  :  A  B ;  then, 
representing  F  A  by  ch,  and  A  B  by  w,  we  have 

Sin.  F2  :  w  ll  sin.  B2  :  ch  =  ~^-JF --  =  sin~^  '  *™  F'    =  Sin7r3 

Sin.  C  :  ch  : :  sin.  A  :  r  +  J  h  =  -^5^-  =  (^7yf  s'f  c- 
then,  by  subtracting  J  h  we  have  r. 

EXAMPLE      OF     COMPUTATION. 

Let  Fr  =  7°  34'  29";  Sw  =  1°  08'  12";  h  =•  4-7  feet;  d  = 
0-41GG;  then, 

900  90o  I80o 

JFV=   70  34' 29"  Sw=    10  08' 12"  C    =         6 o  26' 17" 


900_Fr  =  82025'31"  =  F          9QO  +  Sw  =  91°  08'  12"  =  B  2)l73O33'43" 


910  08' 12"  =  B  90o  A    =       86O46'51"-5 

180°  —  (B  +  F)  =  6°  26'  17"  =  C  Sw  =    1°  08'  12"  F    =       82°  25'  31" • 


1800  oo'  00"  900  —  Sw  =  88°  51'  48"  =  B2          F.,  =         40  21'  20" -5 


TURNOUTS      OUTSIDE      THE  CURVE.  81 

F2  =    4°  2V  20"-5co.  ar.  sin.  =  1-1194836 

C  =6°  26'  17"      co.  ar. sin.  =  0-950282? 

A       =  86°  46'  51"-5      sin.  =  9-9993142 

h  —  d    =4-2834  log.  =  0-6317886 


r  4.  1  h      =  502-19  feet  ........     log.  =  2-7008686 

k       =       2-35 


r        =  499-84 

We  have  thus  found  the  radius  ==  499*84  ;  it  was  intended  as 
a  reverse  of  the  previous  problem,  which  gave  499*725  feet.  The 
difference,  it  will  be  perceived,  is  a  mere  trifle,  and  is  owing  to  the 
loss  of  small  fractions  in  the  frog  angle,  and  by  using  tables  of 
limited  extent. 

(39)  The  next  form  of  a  turnout  which  we  shall  consider,  is 
one  which  shall  turn  out  upon  the  outside  of  a  curve. 

Retaining  the  same  length  of  switch  we  had  in  our  preceding  cal 
culation,  of  course  the  switch  angle  will  remain  the  same.  Then, 
representing  the  switch  angle  by  Sw  ;  the  slide  motion  by  d  ;  the 
radius  of  the  main  track  by  r;  the  radius  of  the  turnout  by  /; 
the  radius  of  the  side  track  by  /';  and  the  line  C  C'  by  a,  we  have 
in  the  small  triangle  C  A  CT,  two  sides,  and  an  included  angle,  viz., 
the  angle  at  A  =  the  supplement  of  the  switch  angle  Sw,  and 
C'  A  =  r;  C  A  =  r,  to  find  the  remaining  side  a,  and  the  angles 
C  and  C'. 

This  problem  has  been  so  often  investigated,  and  is  so  well  under 
stood,  that  I  have  deemed  it  unnecessary  to  give  an  investigation. 
We  however  give  a  formula  in  connection  with  the  investigation 
of  the  turnouts,  so  that  the  computer  is  enabled,  without  being 


USEFUL      FORMULA. 

obliged  to  look  up  at  the  time,  elementary  works  to  supply  the 
deficiency  of  such  papers,  or  to  refresh  his  memory,  where,  per 
chance,  he  may  he  somewhat  in  doubt. 

In  the  following  formula  we  shall  use  the  symbols  by  which  we 
represent  the  triangle  under  consideration. 


Tan.  X  =  ta^H^o-AH(r+d)c..rr  .   and  x  +  i  (180o  _  A)  =  CT  ; 

and  X  —  |  (180°—  A)  ==  C. 

And  sin.  C'  :  r  +  d  :  :  sin.  A  :  a  =  ^±^^-  (40) 

or,  by  way  of  proof,  we  have 

Sin.  a:/   ::  sin.  A  :  a  =  -=^1 

which,  if  our  previous  computations  have  been  correctly  performed, 
the  results  of  this  and  the  preceding  analogy  will  be  alike. 

(40)  Having  solved  the  small  triangle,  we  next  endeavor  to 
find  the  magnitude  of  the  frog  angle,  and  its  relative  position  in 
the  main  track.  For  the  accomplishment  of  which  object,  we  have 
in  the  triangle  C  F  C'  the  three  sides,  to  obtain  their  angles. 

Having  obtained  their  angles,  we  then,  by  the  solution  of  the 
triangle  C  A  F,  obtain  the  chord,  which  we  shall  represent  by  ch,  and 
which  will  give  us  the  distance  of  the  frog  angle  in  the  main  track 
from  the  mouth  of  the  switch  upon  the  outside  rail  of  the  turnout, 
or  from  A  to  F.  We  also  give  the  following  formula,  without  going 
into  a  general  investigation,  using  the  symbols  by  which  we  repre 
sent  the  lines  of  the  present  triangle  under  consideration.  Repre 
senting  the  guage  of  the  track  by  h,  we  have,  in  the  triangle 
CFG',  the  line  C  C'  =  a,  obtained  by  (40  ;)  the  line  C  F  =  r  + 
J  h,  which  we  represent  by  b;  the  line  C'  F  =  /  +  \  h,  which  we 
represent  by  c. 


TURNOUTS      OUTSIDE      THE      CURVE.  83 

If  wo  now  put  p  =  J  (a  -f-  b  -f-  c)  we  have 
Tan.  t  F  =  (Jt.«J^)*;  and  tan.  J  C'  -  (^"^1^-)*; 

and  tan.  I  C^;;^^)*;  (41) 

and  180°  —  F  =  frog  angle  ==  F. 

Substituting  C'2  for  C'  as  obtained  by  formula  (40,)  we  have 
Sin.  i  [180°—  (C'—  C'O]  :  c  ::  sin.  (C' —  C'2)  :  ch  = 

c  .  sin.  (C'—  C'a) 


sin.  i (1800— C'—C'a) 

or,  probably,  in  practice,  the  following  may  be  substituted  with  con 
venience,  viz., 

E  :  2c  : :  sin.  i  (C'  —  C'2)  :  ch  =  sin.  J  (C'  —  C'2)  2c      (42) 

(41)  The  next  step  in  our  investigation  will  be  to  ascertain  the 
reversing  point,  M ;  and  the  terminus  of  the  reverse  curve,  T.  For 
this  purpose,  in  the  triangle  C  C'  C"  we  have  three  sides,  viz.,  the 
side  C  C'  =  a,  from  (40 ;)  side  C'  C"  =  2  r  ;  and  the  side  C  C"  == 
C  T  —  /;  and  C  T  =  r  +  8;  therefore,  C  C"  =  r  +  8  —  /. 

Ecpresenting  by  S  the  distance  between  centre  lines  of  the  main 
and  side  tracks,  and  substituting  b  for  C  C",  and  c  for  C'  C",  we 
obtain  the  angles  by  (41  ;)  and  then,  to  find  A  M,  which  we  repre 
sent  by  ch',  we  have 

E  :  2r  : :  sin.  i  (C'  —  C'«)  :  cli  =  2r  .  sin.  \  (C'"—  C'2)       (43) 

Then,  to  find  the  chord  M  T,  which  we  represent  by  ch" ,  we  have 
B  :  2r'  : :  sin.  i  (180°—  C")  :  ch"=2r.  sin.  i  (180°—  C")  (43) 

To  lay  off  these  chords,  we  place  the  instrument  at  the  centre  of 
the  mouth  of  the  switch,  when  switched,  pointing  in  the  direction 


84  USEFUL     FORMULA. 

parallel  to  the  tangent  of  the  curve  of  the  main  track,  (the  tangent 
to  the  curve  of  the  turnout  will  be  found  to  vary  from  the  tangent 
of  the  main  track  in  amount  equal  to  the  switch  angle ;)  lay  off  an 
angle  towards  the  turnout  side  of  the  road  =  90°  —  Snv  + 
2  (180° — C';)  then,  measure  the  chord  ch  to  M,  the  reversing 
point ;  then,  moving  the  instrument  to  M,  and  pointing  it  to  the 
station  between  the  mouth  end  of  the  switch-bars  just  left,  lay  off 
an  angle  on  the  side  towards  the  main  track  equal  to  180°  — 
i  (180°—  C')  +  i  (180°  —  C";)  then  measuring  the  chord  ch"  to 
the  tangent  point  T. 

We  think  that  further  details  of  the  method  of  locating  the 
curves  need  not  be  here  given,  the  principles  having  been  fully 
explained  in  the  foregoing  pages. 

To  proceed  with  an  example  of  computation.  We  put  r  = 
5729*597  feet;  /  =  499 -72.5  feet;  7*  =  4-7  feet;  <2  =  0-416 
feet ;  5  =  11  feet ;  Sw  =  1°  08'  12"  ;  and  of  course  A  = 

1780  si/  48" 


2)  1°  08'   12" 

00  34'  06"  (40) 


r  +  d  =  5730-013 

r  =     499-725 


(r  +  d)  +  r'  =  6229-738                co.  ar.      log.  6-2055302 

(r  +  d)  «~    r'  =  5230-288  log.  3-7185256 

£  (1800—  A)    =  00  34'  06"  tan.  7-9964947 

X                          =  00  28'  37" -772  tan.  7-9205505 


C'  =1°  02'  43" -772 

C  05'   28" -228 

A  =   178°  5i/  48" 


Proof  1800  QO7  00" -000 


COMPUTATION      OF      ELEMENTS. 


85 


C'   =1°  02'  43"  -772  co.  ar.  sin.  =  1  •  7388259 

r  -f-  d  =  5730  •  013  feet  log.  =  3  •  7581556 

A    =178°  51'  48"  sin.  =  8  •  2974820 

a    =  6229  •  647  feet  log.  =  3  •  7944635 


C  =  00  05'  28" -228  co.  ar.  sin.  =2-7982497 

499-725  feet  log.  =  2  •  6987311 

8-2974820 


3-7944628 


NOTE.    By  a  more  strict  computation,  the  second  analogy  gave 
the  same  results  as  the  first. 


=  5729-597 
=    2-350 
A  =  5731-947  =  b 


r>  =  499-725 

1  h  =      2-350 


r'+  J  h  —  502-075    =  c 


(41) 


a 

=  6229-641 

b 

=  5731-947 

c 

=  502-075 

2^12463-663 

P 

=  6231-831 

log. 

=  3-7946157 

p>-  a 

—    2-190 

log. 

=  0-3404441 

p  -  ft 

=  499-884 

log. 

=  2-6988693 

p  —  c  =  5729-756   log.  =  3-7581362 

p      co.  ar.  log.  =  6-2053843 

p  —  a  co.  ar.  log.  =  9  •  6595559 

p  —  b  log.  =  2-6988693 

p  —  c  log.  =  3-7581362 


2)22-3219457 

F  =  86°  03'  04" -44  tan.  =  11-1609728 
2 


F    =172°  06'  08"' 


p  co.  ar.  log.  =  6-3053843 

p  --  b  co.  ar.  log.  =  7-3011307 

p  —  c  log.  =  3-7581362 

p  —  a  log.  =  0-3404441 


•6050953 


JC'=*  3°  37'  53" -46    tan.  =     8-8025476 
2 


70   is/  46" -92 


7°  53'  51" -12  =  the  frog  angle. 


p  co.  ar.  log.  =  6-2053843 

p  —  c  co.  ar.  log.  =  6-2418638 

p  —  a  log.  =  0-3404441 

p  —  b  log.  =  2-6988693 

2)  15 -48656 15 

C   =  00   19'  02" -09    tan.  =  7-7432807 


F,=s  1720  06'  08" -88 
C'  =  7°  15'  46" -93 
C  =  00  38'  04"- 19 


1800  00'  00" -00 


Proof  by 

adding  angles. 


C   =00 


04" -18 


USEFUL      FORMULA. 


C'  =70  15'    46" -93 

C'2  =1°  02'  43" -77 

2^6°  13'  03" -16 
i  (C'—  C'a)  =  30  06'  31" -58 
T  -\-  \  h  =  502-075  feet 

2- 
ch        =  54-457  feet 


sin.  =  8-7342531 
log.  =  2-7007686 
log.  =  0-3010300 


log.  =  1-7360517 


(42) 


a              =  6229-641 

r  +  <5  —  r'  =  b  =  5240-872 

2  r  =  c        =  999-450 

2^12469-963 

p              =  6234-981   log.  =  3-7948352 

p  —  a         =  5-340   log.  =  0-7275413 

p  —  b          =  994-109   log.  =st  2-9974341 

p  —  c          =  5235-531   log.  =  3-7189608 


p  CO.  ar.  log.  =  6-2051648 

p  —  a  co.  ar.  log.  =  9  •  2724587 

p  —  b  log.  =  2-9974341 

p  —  c  log.  =  3-7189608 


2 j 22 -1940184 

85°  25'  37" -62   tan.   =  11-0970092 
2 


C"  =  170°  51'   15" -24 


p  co.  ar.  log.  =±  6-2051648 

p  —  b  co.  ar.  log.  =  7-0025659 

p  —  c  log.  =  3-7189608 

p  —  a  log.  =  0-7275413 


3°  50'  32" -15  tan. 


C'  =  70  41'  04" -30 


•  6542328 


•8271164 


p  co.  ar.  log.  =  6-2051648 

p  —  c   co.  ar.  log.  =  6-2810392 
p  —  a  =  0' 7275413 


P  — 


=  2-9974341 
2^16-2111794 


43'  50" -21  tan.  =  8-1055897 
2 


10  27'  40" -42 


C"  =  1700  5 1/  15". 25 
C'  =  7°  41'  04" -31 
C  =  1°  27'  40"'43 


1800  00*  00" -00 


$  (1800  _  c")  =  40  34'  22" -37  sin.   B- 9016660 

r  =  499 -725  feet  log.  2 • 698731 1 

2-000  log.  0-3010300 

ch"  =     79 -683  feet  log.  1-9013671 


(43) 


TURNOUT      TO      GIVEN      FROG.  87 

We  have  thus  completed  the  computations  necessary  to  find  all 
the  leading  or  principal  elements  of  a  turnout  upon  the  outside  of 
a  curve  in  the  main  track;  and  whatever  of  unexplained  detail 
may  he  required  can  he  computed  in  the  field,  as  the  computations 
will  he  hoth  short  and  simple. 

Let  us  now  reverse  the  prohlem  hy  supposing  that  we  possess  a 
frog  of  given  dimensions,  and  are  desirous  to  make  it  serve  us  in  a 
turnout  from  the  outside  of  a  curve  in  the  main  track,  whose  radius 
of  course  we  know.  It  will  then  "become  necessary  to  ascertain  a 
radius  for  the  turnout  which  will  he  suited  to,  or  compare  with,  the 
angle  of  the  frog. 

(42)  Without  further  remark  we  will  proceed  to  the  investiga 
tion  of  a  formula.  To  render  this  investigation  plain  to  the  under 
standing,  it  may  he  necessary  to  "become  rather  more  particular  in 
describing  the  figure  or  diagram  upon  which  it  is  hased,  than  it  has 
heretofore  heen  our  custom.  Making  use  of  the  same  notation  of 
the  preceding  prohlem,  as  far  as  applicahle,  we  will  commence  the 
construction  of  the  figure  at  the  point  where  the  angle  of  the  frog 
is  to  be  placed  in  the  outside  rail  of  the  main  track,  viz.,  at  F ; 
from  thence  we  draw  a  line  to  C  =  r  -f~  |  h,  for  the  radius  of  the 
outside  rail  of  the  curve  in  the  main  track,  and  describe  a  portion 
of  the  arc ;  then,  from  the  same  centre,  with  a  radius  =  r  +  |  h  -f- 
d,  describe  the  arc  S  s;  and  with  a  radius  —  r  —  J  h,  describe  the 
inner  rail  of  the  main  curve.  At  F  lay  off  an  angle  from  F  C  = 
the  frog  angle  +  the  switch  angle,  and  in  accordance  therewith 
draw  the  line  F  C"  =  r  —  J  h  +  d;  and  then,  from  the  point  C" 
as  a  centre,  with  a  radius  =  r  -f-  \  h,  describe  the  arc  A  S ;  the 
intersection  of  the  arc  with  the  little  arc  S  s  at  S  will  be  the  place 


TURNOUT      TO      GIVEN      FROG.  89 

of  the  mouth  end  of  the  switch-rails,  when  switched;  then,  draw 
the  radius  from  C  to  S,  and  continue  the  same  indefinitely  on  the 
opposite  side  ;  from  S  draw  the  line  S  C',  making  an  angle  with  the 
continued  radius  =  the  switch  angle  ;  then,  continue  the  line  C"  F 
indefinitely,  and  draw  the  line  F  C',  making  an  angle  with  the  con 
tinued  line  F  C"  =  the  switch  angle  ;  the  intersection  of  the  lines 
S  C'  with  F  C'  at  C'  will  determine  the  length  of  the  radius  of  the 
turnout.  If  we  now  unite  S  C"  and  C  C",  we  shall  have  a  sym 
metrical  figure  containing  two  triangles,  S  C  C"  and  F  C"  C,  which 
are  similar  and  equal.  We  shall  have  also  the  triangles  S  C  F  and 
F  C"  S,  which  are  similar  and  equal. 

(43)  Having  thus  completed  our  figure,  we  commence  our 
investigation  hy  endeavoring  first  to  find  the  angle  S  C  F, 

We  have  the  angle  C  S  C"  =  the  angle  C  F  C",  and  the  angle 
C  F  C"  as  "before  stated  =  the  frog  angle  +  the  switch  angle.  In 
the  triangle  CSC"  we  have  the  angle  S  =  the  frog  -f-  the  switch 
angle  ;  and  the  side  S  C  =  r  —  \h^rd;  the  side  S  C"  =  r  +  i  7a. 
As  we  have  before  stated,  the  angles  S  C  C"  and  F  C"  C  are  equal  ; 
hence,  it  is  obvious  that  the  angle  sought,  viz.,  S  C  F,  is  =  to  the 
difference  between  the  angles  S  C"  C  and  S  C  C";  to  find  which,  we 
have  (r  +  \  li)  +  (r  —  \li  +  d)  :  (r  +  \  Ji)  -  (r  —  \  h  +  d) 
I  :  tan.  J  [180°—  (Fr  +  Sw)~]  :  tan.  X,  and  2  X  =  C,  the  angle 
sought. 


Having  found  the  angle  C,  our  direct  course  would  be  to  find  the 

angles  S  and  F,  and  the  side  S  F  in  the  triangle  C  S  F,  which  could 

readily  be  done  by  formulae  similar  to  the  above,  viz.,  (44  ;)  but, 

believing  the  following  to  be  more  convenient,  we  pass  that  by. 

N 


90  USEFUL      FOEMUL2E. 

We  therefore  have,  in  the  quadrilateral  figure  C  S  C'  F,  the  angle 
at  C  =  2  X  ;  the  angle  at  S  =  180°  +  the  switch  angle;  the 
angle  at  F  =  (180°-—  frog  angle,)  and  the  angle  at  C'  =  the 
explementary  angle,  or  which  shall  make  the  sum  of  all  the 
angles  =  360°.  It  is  now  apparent  that  it  will  he  convenient  to 
represent  hy  the  letters  C  S  C'  F,  the  angles  belonging  to  three 
distinct  figures,  viz.,  the  angles  of  the  quadrilateral  just  named  ; 
the  angles  of  the  triangle  C  S  F,  and  of  the  triangle  C'  S  F.  For 
the  purpose  of  preventing  confusion,  when  we  use  the  letters  to 
denote  an  angle  of  the  quadrilateral,  they  will  not  be  accompanied 
by  any  distinguishing  mark.  When  to  denote  an  angle  in  the 
triangle  C  S  F,  they  will  be  marked  thus,  Ci  Si  Fi  ;  and  when  to 
denote  an  angle  in  the  triangle  C'  S  F,  they  will  be  marked  thus, 
€'2  82  F2.  We  shall  also  denote  the  frog  angle  by  Fr,  and  the 
switch  angle  by  Sw. 

To  proceed  with  the  investigation,  we  have 

i  (180°  —  C'2  )  =  S2  =  F2  ;  and  F  --  F2  =  Ft  ; 

and  S  —  S3=Si. 

Having  thus  obtained  all  the  angles  of  both  triangles,  we  have 

Sin.  Si  :  r  +  |-  h  :  :  sin.  Ci  :  S  F  =  -^±^;.^L.        (45) 

Substituting  c  for  S  F,  we  have  an  equal  expression,  which  we  fre 
quently  use  by  way  of  proof  to  our  work,  viz., 

Sin.  Fi  :  r  —  H  +  d::sin.C,  :  0=^-^^^-    (45) 


We  next  have  sin.  C'  2  :  c  ::   sin.  Fa  or  82    :  r  =  the  radius  of 
the  turnout  sought.  (46) 


TURNOUT      TO      GIVEN      FROG.  91 

EXAMPLE     OF     COMPUTATION. 

We  will  suppose  r  =  5729  -597  feet;  Fr=  T  53'  51" -12;  Sw  = 
1°  08'  12",  to  find  the  radius  of  turnout  /. 


Fr 

Sw 
Fr  -f  Sw 

=    7°  53'  51"'  12 
=    10  08'  12" 

=    90  02'  03"-  12 

2)l70°  57'  56"' 88 
i  (180°  —  [pr  +  SW])          ^850  28'  58"' 44 


r  +  4  h  5731-947 

r  _  £  h  +  rf  5727-663 


(r  _|_  i  fc)  _|_  (r  _  j  A  +  d)    =  11459-610  co.  ar.    log.  =  5-9408302 

(r  +  J-  A)   co  (r  —  i  A  +  d)    =          4-284  log.   =  0*6318495 

£  (1800  — [Fr  +  Su>])=  —     850  28'  58" -44  tan.  =  1-1023618 


X  =      00  16'  16" -03  tan.  =  7-6750415 

(44) 


C   =  2X  =  00  32'  32"-06 

(1800  _  pr)  =  F  =  1720  06'  08" -88 

18QO  4.  Sw  =  S  s=  1810  08'   12"-00 

C'  —  60   13'  07" -06 


3600   0(X   00" -00 


1800  00'  00" -00 
C'a  =60   13'   07" -06 


2  j  1730  46'  52" -94 

S2   =  F2   =  i  (1800—  c'a)        =     860  53'  26"-47          S2   =     86°  53'  26"-47 
F  =   1720  06'  08" '88  S    =181°  08'  12" 


F!  =850   12'  42" -41  Sx   =     94O   14'  45" -53 

Si  =940   14'  45" -53 

2X  =  C,  =       00  32'  32"-06 


180°   00'  00" -00 


S,  =     940  14'  45"-53    co.  ar.    sin.  =  0-0011936 

r  +  i  h  =     5731-947  feet  log.  =  3-7583021 

C,  =       QO  32'  32" -06  sin.   =  7-9760615 


c  log.  ==  1-7355572 


[Fia.     16.] 


TURNOUT  FROM      INSIDE      OF  CURVE.                  93 

Fi  =850  12'  42" -41    co.  ar.    sin.  =  0-0015183 

r  —  £  h  +  d  =      5729-663  feet  log.  =  3-7579775 

G!  =        00  32'  32" -06  sin.  =  7-9760615 


c  =  log.   =  1-7355573 

C'a  =60   13'  07" -06  co.  ar.    sin.   =  0-9652810 

F2  =86°  53'  26" -47  sin.   =  9-9993601 

r'  =     501-41  feet  log.  =  2-7001984    (44) 

We  have  thus  found  /  =  501  '41  feet.  It  was  intended  as  a  reverse 
of  the  previous  case;  there  we  assumed  r  =  499 '725  ;  the  differ 
ence  is  a  trifle,  being  only  1-685  feet,  which  is  not  astonishing 
when  we  consider  the  acuteness  of  the  angles  we  have  to  use  in 
some  of  the  triangles. 

I  ought  not  to  close  my  remarks  without  an  acknowledgement 
of  my  indebtedness  to  Mr.  Percival,  of  Sandwich,  for  the  man 
ner  of  constructing  the  figure  which  has  led  us  to  the  foregoing 
investigation. 

(44)  We  next  examine  a  turnout  from  the  inside  of  a  curve 
in  the  main  track. 

Eetaining  our  former  notation  as  far  as  practicable,  we  have  in 
the  triangle  S  C  C',  the  line  S  C  =  r  —  d;  S  C'  =  /;  angle  S  == 
switch  angle  to  find  the  side  C'  C  (which  we  denote  by  «,)  and  the 
angles  C'  and  C  ;  wherefore,  (r  —  d)  +  /  I  (r  —  d)  ^  r'  : :  tan.  J 

(180°  —  S)   :   tail.    X  =      t»"-H180°^S).|r-d)  ~  rO      and    j     ^QQO  _  g) 

+  X  =  C';  and  \  (180°  —  S)  —  X  =  C. 


Then,  sin.  C'  :   r  —  d  ::  sin.  S   :   s  =    r~^sg;s    ;  or,  we  have 


sin.  C  :  /  : :  sin.  S    :  s  =  — ^f-  (47) 

Having  found  s,  we  have  in  the  triangle  C  C'  C"  the  side  C  C 


94  USEFUL      FORMULA. 

=  s,  as  found  above;  the  side  C'  C"  =  2/;  and  the  side  C  C"  = 
(r  —  S)  +  /  to  find  the  angles,  which  we  do  by  formula  (41) 

Having  thus  found  the  angles  required,  we  will  denote  the 
triangle  SCO'  No.  1,  and  represent  the  angles  in  said  triangle 
by  Si,  Ci,  C'i  ;  and  the  triangle  C  C'  C"  No.  2,  and  represent 
the  angles  by  Cs,  C'  2  ,  C"  2  ;  and  the  triangle  S  C'  M  No.  3,  and 
represent  the  angles  in  said  triangle  by  Sa,  C's,  M3,  and  the 
triangle  C"  M  T  No.  4,  and  shall  accompany  the  letters  denoting 
the  angles  by  4  ;  and  so  on  of  such  other  triangles  as  may  enter 
into  our  investigation  in  the  order  they  are  presented. 

We  will  now  proceed  to  find  the  chord  S  M,  which  we  shall 
denote  by  ch  3.  In  triangle  No.  3,  we  have  C'  a  =  C'  i  —  C'  2  ;  and  J 
(180°—  C3)  =  S3  =  M3;  then  will 

Sin.  S3  :  r  :  :  sin.  C'  3  :  chz=   r/sin-^_  (48) 

sin.  Si 

In  triangle  C"  M  T  =  No.  4,  we  have  C"  4  =  C"  2  ;  and  \ 
(180°  —  C"  4)  =  M4  =  T4;  then, 

Sin.  M4  :  r  ::  sin.  C"4  :  ch*  =  ^sin-  c//*  (49) 

sin.  M4 

Then,  putting  CFG'  for  No.  5,  we  have  C'  F  =  r  +  }  h;  C  F 
=  r  —  J  Ji;  and  C  C'  as  found  in  No.  1  ,  (which  we  called  s,)  to  find 
the  angles.  See  formula  (41) 

Having  found  the  angles,  the  angle  C'  F  C,  or  Fs,will  =  the 
frog  angle.  Then,  to  ascertain  the  chord  S  F,  we  have  in  the  triangle 
C'F  S  =  No.  G,C'z  —  C5  =  C6;  andi(180  —  C'e)  —  S6  =  Fc,and 


sin.  S6  :  /  +  i  h  :  :  sin.  CT  .  :  ch,  =  0"  +  *  »)  .  "in.  c'6  (50) 

sin.  S6 

which  represents  the  distance  from  the  mouth  of  the  switch  of  the 
outside  rail  of  the  turnout  to  the  frog  angle. 


TURNOUT      FROM      INSIDE      OF      CURVE. 


95 


EXAMPLE      OF     CALCULATION. 

Let  r==  5729 -597  feet;/ 499 -725  feet;  A  ==4 -7  feet; 
feet;  5=11  feet;  Si ,  or  switch  angle,  =  1°  08'  12". 


r'  —  d 
r' 


=  5729-181 
=     499-725 


(r  _  d)  -f  r'  —     6228-906 
(r  —  d)   c~   r'  =     5229-456 
£  (18QO  _  Si)  =    890  25'  54" 
X  =890   19'  23" 


180°  00'  00" 
Si  =  1°  08'   12" 

2^1780  51'  48" 
J  (1800—  §i)  =  890  25'  54" 
co.  ar.    log.   =  6-2055882 
log.   =  3-7184566 
tan.  =  2-0035053 


tan.  =  1-9275501 


C\  =  178°  45'   17" 

Ca  =00  06'  31" 

S1  =10  08'   12" 

1800  00'  00" 

Ci          =1780  45/  17"  Co.  ar.   sin.  =  1-6628906 

r  —  d  =  5729-181  feet  log.  =  3-7580926 

Si          =10  08'  12"  sin.  =  8-2974820 

s,           =5229-532  log.  =  3-7184617 


(47) 


C,   =  00  06'  31"    co.  ar.    sin.  =  2-7222486 

r'     =  499-725  feet  log.  =  2-6987311 

sin.  =  8-2974820 

log.  =  3-7184617 


To  find  the  elements  of  triangle  No.  2,  we  have 


—  r  —  5  +  r'  =  6218-322 


c  —  2r' 

=  5229-532 
=  999-450 

2)l2447-304 

P 

=  6223-652 

log.  = 

3-7940453 

p  —  a 

=    5-330 

log.  = 

0-7267272 

p  -  b 

=  994-120 

log.  = 

2-9974388 

p  —  c 

=  5224-202 

log.  = 

3-7180200 

P 

co.  ar.   log.  = 

6-2059547 

P 

co.  ar.   log. 

=  6-2059547 

p  —  a 

co.  ar.   log.  = 

9-2732728 

p  -  b 

co.  ar.   log. 

=  7-0025612 

p  -  b 

log.  = 

2-9974388 

p  —  c 

log. 

=  3-7180200 

p  —  c 

log.  = 

3-7180200 

p  —  a 

log. 

=  0-7267272 

2)22-1946863 
85°  25'   50" -22    tan.  =   11-0973431 

2 


2) 17 -6532631 
30  50'   16" -79    tan.   =  8-8266315 
2 


C'a  =   1700  5  i>  40" -44 


C"2   = 


33"  -58 


96  USEFUL     FORMULA. 

p  co.  ar.       log.  i=  6-2059547 

p  —  c  co.  ar.       log.   =  6-2819800 
p  —  a  log.  =  0-7267272 

p   —  b  log.   =  2-9974388 


2J16-2121007 
00  43'  53" 
2 


C2       =  10  27'  46" 

C'a  —  1700  5i/  40" -45 

C"2  =      7°  40'  33" -58 

C2  =10  27'  46" -00 


Proof  1800  otX  00" -00 

In  triangle  No.  3,  we  have 

C',  =   1780  45/   i7'/ 


C'3  =70   53'  36" -56 


2  )  1720  06'  23" -44 

i  (180  —  C',)  =  M3=     860  03'   11" -72  co.  ar.     sin.  =  0-0010312 

r'                                  =  499-725  feet  log.  =  2-6987311 

C',                                  =       70  53'  36" -56  sin.  =  9-1377717 

ch3                               =     68-791  feet  log.  =  1-8375340 

In  triangle  No.  4,  we  have 

C"2    =  C"4  =    70  4(X  33" -58 


J  (1800—  C"4)  =  M4  =860  09'  43". 21  co.  ar.    sin.   =  0-0009751 

r'                                      =499-725  feet  log.  =  2-6987311 

C4                                      =7°  40'  33" -58  sin.  =  9-1257121 

ch4                                  =    66-899  feet  log.  =  1-8254183 

In  triangle  No.  5,  we  have  C   F  =  r —  5  h;  C  C"  =  s 
5229-565  ;  C'  F  =  /  +  \  h,  to  find  the  angles.     Let 

a  =  r  —  £  h   =  5727-247 
b  =  r'  +  i ;  A  =  502-075 


log.  =  3-7581123 
log.  =  0-3416323 
log.  =  3-7182831 

log.  =  2-i 


2)ll458-887 

P 

=  5729-443 

p  —  a 

=        2-196 

p  -b 

=  5227-368 

p  —  c 

=    499-878 

TUENOUT   FROM   INSIDE   OF   CURVE. 


97 


p  co.  ar.    log.  =  6-2418877 

p  —  a  co.  ar.    log.  =  9-6583677 
p-b 
p  —  c 

86°  oi'  50" -13 


C'5  =  172°  03'  40" -26 


p  co.  ar.    log.  =  6-2418877 

p  —  c  co.  ar.    log.  =  7-3011359 

p  —  a  log.  =  0-3416323 

p  —  b  log.  =  3-7182831 


2^17-6029390 

30  37'  21" -13        tan.  =  8-8014695 
2 


F,      =70  14'  42"  -26    Frog  angle. 


In  triangle  No.  6,  we  have 


C's 


p  co.  ar.    log.  =  6-2418877 

p  —  b   co.  ar.    log.  =  6-2817169 


log.  =  3-7182831 
log.  =  2  '  6988641 

p  —  c 
p  —  a 

00  20'  48"  -73 
2 

log.  =  2-6988641 
log.  =  0-3416323 

2)22-3174026 

2^15  -5641010 

tan.  =  11-1587013 

tan.  =  7.7820505 

C.     =  00  41'  37"  -46 

Cs      1720  Q3/  40" -26 
FK         70  14'  42" -27 


1800  00'  00" -00 


=     1780  45'  17" 
=      1720  05'  24" -65 


=          60  39'  52" -35 


211730  20'  07" -65 


=  J  (1800—  C'6)      =       860  40'  03"- 


S6  =860  40'  03" -82 

r1  +  J  h  =  502  •  075  feet 

C6  =60  39'  52" -35 

ch6  =    58-233  feet 


co.  ar.  sin.  =  0-0007349 
log.  =  2-7007686 
sin.  =  9-0646679 


log.  =  1-7661714 


the  distance  from  the  mouth  of  the  switch  on  the  outside  of  the 
turnout  to  the  frog  angle  in  the  main  track. 


TURNOUT      FROM      INSIDE      OF      CURVE.  99 

(45)  Having  thus  completed  our  investigation  of  the  problem 
direct,  we  will  now  examine  it  reversed,  by  supposing  the  radius  of 
the  main  track  given,  as  before,  viz.,  r  =  5729 '597;  h  =  47  feet ; 
d  =  0  -416  feet ;  and  the  frog  angle  Fr  =  7°  14'  42".  27  ;  the  switch 
angle,  Sw  =  1°  08'  12";  to  find  the  radius  of  the  turnout  =  /,  and 
the  position  of  the  frog. 

Draw  the  lines  ee,  cc  representing  the  outer  and  inner  rail  of  the 
main  track,  and  the  dotted  line  //,  corresponding  to  the  switching 
of  the  outer  rail ;  then,  draw  the  radius  F  C  ;  then,  from  F  draw  the 
line  F  C'  indefinitely,  making  an  angle  with  F  C  =  the  frog  angle ; 
then,  from  F  draw  the  line  F  C",  making  an  angle  with  F  C  =  the 
switch  angle  -f-  the  frog  angle,  and  equal  in  length  to  the  radius 
of  the  dotted  line  =  C  G;  then,  with  a  radius  =  C  F  and  with  C" 
as  a  centre,  draw  the  angular  dotted  line  at  S,  and  this  dotted  line 
will  intersect //at  the  mouth  of  the  switch.  From  this  intersection 
draw  the  radius  S  C  ;  then,  draw  the  line  S  C',  making  an  angle  with 
S  C  =  the  switch  angle,  and  the  lines  S  C'  and  F  C'  will  intersect 
each  other  at  the  centre  of  the  curve  of  the  turnout,  viz.,  at  C'; 
then,  with  a  dotted  line,  join  S  C"  and  C"  C,  which  will  complete  our 
diagram. 

If  we  now  examine  our  diagram,  we  shall  find  it  to  contain  two 
equal  and  similar  triangles,  viz.,  FCC"  and  S  C"  C  with  the 
angles  F  C"  C  =  S  C  C";  then  it  will  be  apparent  that  the  angle 
S  C  F  will  be  equal  to  the  difference  between  the  angles  F  C  C"  and 
F  C"  C.  Having  thus  shown  the  relative  magnitude  of  the  angles 
last  named,  we  will  noAv  proceed  to  find  the  angle  S  C  F.  In  the 
triangle  F  C  C"  we  have  the  angle  at  F  =  the  frog  angle  -f-  the* 
switch  angle;  the  side  F  C  =  r  —  }  li ;  the  side  F  C"  =r+  (i  h 


100  USEFUL      FOKMUL^E. 

-  d;)  then,  (r  +  \  li  —  d)  +  (r  —  J  h)  :  (r  +  \  h  —  d)  ^  (r  — 
4  h)  : :  tan.  4  [180  —  (Fr  +  Sw)]  :  tan.  X ;  that  is, 

~  *h  >  • tan-  *  t180  —  (fr  +  SuQ]  /*1\ 

d)        r-iA) 


And  2  X  =  F  C  S,  (51,)  which  we  shall  hereafter  represent  by  €2. 

In  the  above  notation,  we  have  represented  the  frog  angle  by  Fr, 
and  the  switch  angle  by  Sw. 

We  now  have  the  triangle  F  C  S,  which  we  shall  hereafter 
denominate  No.  2  ;  the  angle  C  =  C  2 ,  as  found  above ;  the  line  C  F 
=  r  —  4  h>  the  line  CS  =  r  +  iA  —  d,  to  find  the  angles  at  F 
and  S,  which  we  shall  denote  thus,  by  Fa  and  82;  then,  (r  -\~  5  h 
-  d)  +  (r  —  4  h)  :  (r  +  %h  —  d)c~  (r  —  4  h)  :  I  tan.  4  (180 
—  C2)  :  tan.  X2  and  4  (180—  C2)  +  X2=F2,  and  J  (180  - 

n    \  V2  __  g  (52^ 

Then,  sin.  F2  :    r  ~\-  \  li  — d  :  I   sin.  C2  :  S  F,  which  we  shall 
denote  by  ch,  or  sin.  82  :  r  —  J  7i  .' :  sin.  €2  :  ch.  (53) 

We  then  have  in  the  triangle  S  F  C',  (which  we  denominate  No.  3, 
and  mark  the  letters  denoting  the  angles  accordingly,)  the  line  S  F 
=  ch,  found  above;  the  angle  83  =  (82+  Sw;)  the  angle  F3  = 
(F2—  Fr)  =  (82+  Sw,)  and  the  angle  C'3  =  180—  [(F2  —  Fr) 
+  (82  +  Sw;)]  and  C'  3  :  ch  :  l  Fs  :  /  +  4  h  (54) 

Then,  by  subtracting  J  h,  we  have  /. 

Having  thus  obtained  our  formula,  we  now  give  an  example  of 
calculation. 


TURNOUT      FROM      INSIDE      OF      CURVE. 


101 


To  find  /  from  the  frog  angle, 


Fr 

Sw 

Fr  +  Sw 

*  [1800  —  (Fr  +  Sw;)] 

—  d  =    5731-531 

=    5727-247 


=        70  14' 
=         1°  08' 


42" -27 
12"- 


=         8°  22'  54"- 27 


2)171°  37'  05"- 73 


—       85°  48'  32"' 86 


Sum  =  11458-778 

Difference  =         4-284 

i  [1800—  (Fr  -f  SuOl     =850  48'  32"- 46 
X  =    00  17'  32"- 38 

2 


log.  =  5-9408619 
log.  =  0-6318495 
tan.  =  1-1350427 


tan.  =  7-7077541 


(1800—  ca) 


=  00  35'  04" -76 
2)  1790  24'  55"- 24 
=  [890  42'  27"- 62 


i  (1800—  C2) 


=    5731-531 

=    5727-247 

11458 • 778 

4-284 

=   890  42'  27"- 62 
=     40  11'  27"- 14 


log.  =  5-9408619 
log.  =  0-6318495 
tan.  =  2-2922459 


tan.  =  8-8649573 


Fs  =930  53'  54"- 76 

Sa  =   850  31'  00"- 48 

C2  =00°  35'  04"- 76 

1800  00'  00"- 00 

=  930  53'  54"-76  co.  ar.  sin.  =  0-0010062        S2  = 

i  — d  =  5731-531  feet  log.  =  3-7582707        r  —  £fc  = 

C2  =    0°  35'  04"-76  sin.  =  8-0087699 

cAa  =     58-620  feet  log.  =  1-7680468 


850  31'  oo"-48  co.  ar.  sin.  =  0-00133C9 

5727-247  feet  log.  =  3-7579458 

sin.  =  8-0087699 

log.  =  1-7680468 


In  triangle  No.  3,  we  have 


Fr 
F3 


=  930  53'  54"- 76 

_  70  i4/  42"- 27 

=  860  39'  12"- 49 

=  860  sg/  12"- 48 


52  =850  31'  00"- 48 
Siv     =      1°  08'  12" 

53  =860  sg/  12"- 48 


1730  is'  24" -97 
18QO  _  (F,  4-  S3)     =      60  41'  35"- 03 


[Fia.     18.] 


TURNOUT      FROM      INSIDE      OF      CURVE.  103 

By  the  problem,  F3  and  S    should  be  equal.     Then, 

1800—  -  (F,  +  S3)    =       6°  41'  35"-03     co.  ar.  sin.  =  0-9334859 

cht  log.  =  1-7680467 

F3  =860  39'  12".  48  sin.  =  9-9992587 

r'  +  £  A  =    502-101 

}  &  =        2-350 


r1  =    499-751 

It  was  expected  the  radius  would  be  found  to  be  499*725.  The 
error  only  amounted  to  '026  feet,  or  a  little  more  than  one  fourth 
of  an  inch. 

(46)  The  following  problem  has  frequently  presented  itself  in 
the  practice  of  the  writer,  viz.,  the  situation  of  a  turning  table 
with  respect  to  the  main  track,  (the  radius  of  the  turnout  curve 
being  given  to  find  the  relative  situation  of  the  switch,)  and  such 
additional  elements  as  will  be  required  to  locate  the  turnout. 

To  explain  :  we  have  in  several  instances  found  it  necessary  so  to 
place  a  turntable  by  the  side  of  the  railroad  track,  that  a  building 
erected  over  it  might  answer  the  purpose  of  shielding  the  table 
from  the  weather,  and  the  engine  during  the  night,  occasionally,  if 
not  constantly  ;  and  also,  to  afford  a  convenient  situation  for  a  water 
tank  to  distribute  water  to  the  engines  when  upon  the  main  track, 
and  when  sheltered  upon  the  turntable. 

The  ruling  principles  which  govern  in  this  matter  may  be  stated 
thus: 

First,  The  proper  distance  between  the  centre  of  the  table  and  the 

main  track. 
Second,  A  suitable  amount  of  straight  track  to  guide  the  engine 


104  USEFUL      FORMULA. 

steadily  upon  the  table.  (A  turning  table  should  never  be  directly 
connected  with  a  curve,  as  the  engine  will  have  a  tendency  to  force 
it  out  of  place.) 

Third,  The  remainder  of  the  track  to  be  united  to  a  curve  of  fixed 
radius,  which  shall  just  connect  the  straight  track  adjoining  the 
table  with  the  main  track. 

INVESTIGATION.  Kepresenting  the  mouth  of  the  switch  by  S ;  the 
length  of  the  switch  by  s,  and  the  switch  angle  by  $iv;  the  centre 
of  the  table  by  O  ;  the  point  where  the  curve  unites  with  the  straight 
line  adjoining  the  table  by  T  ;  the  centre  of  the  curve  by  C ;  the 
radius  by  /  ;  the  point  on  the  main  track  where,  a  line  being  drawn 
therefrom  to  the  centre  of  the  table  shall  form  a  right  angle  with  the 
centre  line  of  said  track,  by  Q ;  we  shall  then  have  in  the  triangle, 
Sw  C  S,  for  finding  the  line  Siv  C,  which  line  we  represent  by  q, 

R  :  /  -|-  d  •  •  cos-  Sw  :  q  =  (r  -f-  d)  .  cos.  Sw  (55) 

And  in  the  triangle  COT,  (by  assuming  C  O  as  radius,)  we  have 
this  analogy : 

/  :  T  0  : :  (C  O  =  radius,)  :  tan.  C2  =  -r^-          (56) 
And  therefore,  cos.  C2  :  r  : :  R  :  C  0  =  — ^  (57) 

Representing  the  distance  of  the  centre  of  the  turning  table 
from  the  main  track  (viz.,  Q  O)  by  n,  and  the  line  C  O  by^>,  we 
then  draw  the  line  O  A,  parallel  with  the  centre  line  of  main  track, 
and  in  the  triangle  A  O  C  \vc  have 

p:*R::q  —  n:  cos.  C  =  -^  (58) 

If  we  now  deduct  from  the  angle  C,  the  angles  Sw  and  €2,  we 
shall  have  left  the  angle  C  of  the  triangle  S  C  T,  which  we  repre 
sent  by  Cs,  then  will  J  (180  —  Cs)  =  S  =  T ;  and 

Sin.  T  :  r   : :  sin.  C3  :  S  T  =  J^~^2-  (59) 


TURNING   TABLE   AND  MAIN   TRACK.      105 

To  give  an  example  of  calculation,  we  will  assume  ft  =  3G'4 

feet;  /  =  499 -725  feet;  s=  21  feet;  T  O  =  60  feet ;  d  =  0" 
feet;  Sw  =  1°  08'  12".     Then, 

r1  -j-  d                =  500-141  log.  =r  2-6990925 

Sw                        ==   1°  08'   12"  cos.  =  9-9999145 

q                            =  500-043  log.   =  2 -6990070 

TO                  =60  feet  log.  =  1-7781513 

r'                         =     499-725  co.  ar.     log.   =  7-3012689 


C4  =6°  50'  47" -41  tan.  =  9-0794202 

C2  =60  507  47" -41       co.  ar.    cos.  =  0-0031080 

r' 

P 

Sw  =   10  08'   12"          q  —  n 
C2  =  60  50'  47" -41    C 


=     499-725 
=     503-314 
=     463-643 
=  220  54/  02"  -78 

log.   =  2-6987311 

log.   =  2-7018391 
log.  =  2-6661837 

cos.  =  9-9643446 

7°  58'  59"-41    (Sw  +  C,)        =     70  58'  59" -41 
C3  =140  55/  03" -37 


2^1650  04'  56" -63 

T  =  £  (180°—  C3)   =  82°  32'  28" -31  co.  ar.    sin.  =  0-0036904 

r>                          =  499-725  log.  =  2-6987311 

C,                        =140  55/   03" -37  sin.  =  9-4106586 


S  T  =  129-742  feet  log.  =  2-1130801 

Having  thus  ascertained  the  elements  deemed  necessary,  before 
we  commence  the  business  of  location,  we  will  now  proceed  to 
describe  the  operations  necessary  to  execute  the  work. 

(47)  An  examination  of  the  figure  will  render  it  apparent  that 
taking  from  the  complement  of  Cu  the  complement  of  C  will  leave 
the  angle  A  O  T.  Having  thus  obtained  A  O  T,  we  place  our 
instrument  at  O  and  lay  off  said  angle  from  A,  and  measure  the 


106  USEFUL      FORMULAE. 

distance  O  T  to  T ;  then,  moving  the  instrument  to  T,  and  point 
ing  it  to  O,  we  lay  off  the  angle  O  T  S  ==  (90°  +  S)  or  (90°  +  T) 
and  measure  the  distance  T  S  to  S,  the  place  of  the  mouth  of  the 
switch ;  and  if  the  work  has  heen  correctly  prepared,  we  shall  ho 
the  distance  d  from  the  centre  line  of  the  main  track,  upon  the 
side  towards  the  turnout  curve.  The  curve  may  now  he  further 
marked  by  deflections,  agreeably  to  directions  given  in  the  foregoing 
pages. 

(48)  Having  thus  obtained  the  formula  for  computing  the 
elements  of  a  turnout  to  a  turntable,  with  a  given  amount  of 
straight  line  to  guide  the  engine,  it  will  frequently  be  found  con 
venient  to  have  a  formula  to  lay  out  a  track  to  a  turntable  from  an 
existing  joint  in  the  rails  of  the  main  track,  with  a  fixed  radius, 
and  a  fixed  position  for  the  table.  This  method  of  proceeding  will 
save  the  trouble  of  cutting  rails,  and  making  unnecessary  joints  in 
the  main  track ;  and  another  consideration  will  be  that  of  affording 
side  track  room  for  cars  to  stand  upon. 

In  the  following  investigation  we  shall  preserve  the  notation  of 
the  preceding  formula,  as  far  as  applicable. 

Making  5  =  the  distance  S  Q,  (as  measured,)  and  a  ==  S  O,  we 
have,  by  taking  a  as  radius,  the  following  analogy : 

5  :  n  —  d  :  I  (rad.)   :    tan.  S  =  -^-7—  (CO) 

wherein  S  will  be  equal  to  the  angle  at  S  in  the  triangle  Q  S  O  ; 
then,  cos.  S  :  *  ::  B  :  a=  C0838  (61) 

then,  representing  the  angle  S  in  the  triangle  C  S  O  by  82,  we 
have  1)0°  +  Sw  —  S  =  Sa,  and 
r  +  alrf  <~a  : :  tan.  \  (180°  —  S2)  :  tan.  X  =  ^L^A 


TURNING   TABLE   AND   MAIN   TRACK.      107 

and  X  +  i  (180°—  82)  =  the  angle  0  ;  and  X  —  i  (180°—  Sa) 
=  the  angle  C  (62) 

We  then  have  sin.  O  '•  r    : :  sin.  82  :  C  O  ;  or,  sin.  C  :  a  1 1  sin. 
S2  :  C  O  ;  then,  putting p  =  C  O,  we  have 

p  :  E  : :  /  :  cos.  C2  =  ^-  (63) 

And  cos.  C2  :  /  : :  sin.  C2  :  T  O  =  tan.  C2  /  (64) 

Then,  deducting  Ca  from  C  leaves  C3  =  the  angle  C  in  the  triangle 
S  C  T ;  and  \  (180°  —  C3)  =  the  angle  S  ==  the  angle  T  ;  and 

Sin.  T  :  /  : :  sin.  Ca :  S  T  =  ^^-  (65) 

To  give  an  example  of  calculation,  we  assume  n  =  36*4  ;  /  = 
499-725;  d  =  0-416;  Sw  =  1°  08'  12";  5  =  200  feet. 


&               = 

200  feet                  co.  ar.    log.  =  7-6989700 

900  00'  00"  -00 

n  —  d            = 

35-984                                    log.   =   1-5561094 

10  08'   12" 

S                       = 

100   n'  58"  -32                     tan.   —  9-2550794 

90°  +  Su>=  91°  OS'   12" 

S                  =  100   n'  58"  -32 

S                       = 

co.  ar.     cos.   =  0-0069179 

S2                 =800  56'  13"  -68 

S                     = 

200  feet                                  log.   =  2-3010300 

2(990  03'  46"  -32 

a                      = 

203-211  feet                          log.   =  2-3079479    i 

(180°—  S2)=  490  31'  53"  -16 

(GO) 

r' 

499  -725  feet 

a                     = 

203-211 

Sum                = 

702-936                  co.  ar.    log.  =  7-1530842 

Difference       = 

296-514                                  log.   =  2-4720452 

i(18(P-Sa)  = 

49°  31'  53"  -16                     tan.   =  0-0689836 

X                      = 

26°   18'  34"-57                     tan.   =  9-6941130 

O                = 

750  50'  27"  -73 

C 

230   13'   18"  -59 

S, 

80°  56'   13"  -68 

180°   00'   00" -00 


108  USEFUL      FORMULAE. 

©         =75°  50'  27" -73  co.  ar.  sin.  =  0-0133981       C    =  23°  13'  18" -59  co.  ar.  sin.  =  0-4041819 


r'          =  499  -725  feet 

log.  =  2-6987311       a    =203  -211  feet 

log.  =  2  •  3079472 

S.,         =800  56'  13".  68 

sin.  =  9-9945442 

sin.  =  9-9945442 

P          — 

log.  =  2-7066734 

log.  =  2-7066733 

r'          = 

log.  =  2-6987311 

=   10°  55'  27" -50  cos.  =  9-9920577  C    =23°  13'  18" -59  (63) 

C2  =  100  55/  27//.  50 


=  100  55/  27"-50  tan.  =  9-2855789  C3  =  12°  17'  51" -09 


r'  =  log.  =  2-6987311  2J167O  42'  08"-91 

TQ  =  96-452  feet  log.  =  1-9843100    i  (180°—  C3)  =  83O  51'  04"-45  (64) 

T  =  830  51'  04" -45  co.  ar.  sin.  =  0-0025057 

r1  =499 -725  feet  log.  =  2-6987311 

C3  =   120  i7>  si'/- 09  sin.  =  9-3283555 


S  T      =   107-051  feet  log.  =  2-0295923 

Having  thus  ascertained  the  elements  of  the  turnout,  it  remains 
to  describe  the  method  of  locating  or  marking  the  same  upon  the  field. 

We  have  found,  formula  (63,)  Ca  =  10°  55'  27" '5,  the  comple 
ment  to  which  =  79°  04'  32" -5  =  the  angle  COT;  we  have  also 
found,  formula  (62,)  O  =  75°  50'  27" -73  =  the  angle  COS. 

We  now  place  our  instrument  at  O,  and  lay  off  from  S  the  dif 
ference  between  79°  04'  32"  -5  and  75°  50'  27" -73  =  3°  14'  04" -77, 
and  measure  from  O  to  T  96 -452  feet ;  we  then  move  the  instru 
ment  to  T,  and  lay  off  the  angle  S  T  O  =  J  (180°  —  C«)  +  90° 
that  is  =  90°  +  T,  as  found  above  (65)  =  90°  +  83°  51'  04" -45 
=  173°  51'  04"-45,  and  measure  107  -051  feet  to  S;  and  if  the 
field  work  and  computations  have  been  correctly  performed,  the 
point  s  will  be  found  directly  between  the  joints  in  the  rails  in  the 
main  track,  and  0'41(>  feet  from  the  centre  line  on  the  side  of  the 
turnout.  The  curve  may  then  be  further  marked  by  deflections,  as 
heretofore  explained. 


TRACK      OVER      WATER.  109 

(49)  It  is  not  an  unfrequent  occurrence  for  an  engineer  to  be 
required  to  locate  railroad  and  other  curves  in  situations  inaccessible 
to  the  making  of  measurements  in  the  common  or  ordinary  methods. 
These  cases  occur  where  railroads  are  located  across  bays  and 
inlets  of  the  ocean  or  lakes,  and  across  rivers  or  estuaries,  etc.,  etc. 
We  know  of  no  better  method  of  managing  this  matter  than  by 
projecting  a  system  of  triangles  from  a  well-selected  base  to  points 
desirable  to  mark  or  permanently  fix.0  The  calculations  necessary 
for  the  arrangement  of  such  a  series  of  triangles,  when  connected 
with  the  choice  of  the  location  of  the  curve,  and  the  determination 
of  the  necessary  elements  to  carry  forward  the  whole  work  with 
accuracy  and  convenience,  may,  in  some  instances,  be  too  compli 
cated  for  the  invention  of  the  young  and  inexperienced  engineer. 
To  aid  such  in  the  performance  of  their  task  is  the  object  of  the 
present  article. 

We  have  chosen  as  an  example,  an  imaginary  river  of  some  250 
feet  wide ;  but,  before  we  proceed  with  the  investigations,  let  us 
suppose  the  straight  or  tangent  lines  upon  both  sides  of  the  river  to 
have  been  located,  and  sufficiently  marked  to  show  their  relative 
bearings.  Our  first  operation,  then,  is  to  select  such  a  situation  as 
may  be  thought,  upon  a  thorough  examination  of  the  whole  subject, 
the  most  desirable  for  the  location  of  the  curve.  The  point  we 
have  selected  will  be  seen  in  Fig.  19,  marked  1 ;  and  in  Fig.  20, 
marked  as  station  10  of  the  railroad  location. 

Having  determined  on  this  point,  we  commence  the  survey  by 
running  a  line  from  it  to  a  point  in  the  tangent  line  upon  the  same 

*  Care  should  be  taken  to  so  select  the  termini  of  the  base,  that  the  lines  projected  therefrom 
should  intersect  with  each  other  at  the  points  to  be  located  as  near  at  right  angles  as  they  well 
can. 


[Fio.     19.] 

A 


TRACK      OVER      WATER.  Ill 

side  of  the  river,  marked  0  on  the  diagram.  Then,  by  considering 
the  tangent  line  from  0  to  A  to  hear  due  east,  (whatever  may  he  its 
direction,)  we  measure  the  angle  A  0  1,  hy  which  we  determine  the 
line  from  0  to  1  to  hear  S.E.  75°  08'  35" -37,  and  hy  measurement 
we  find  the  distance  152*5  feet.  We  then  place  a  signal  at  station 
2,  and  hy  a  triangulation  we  ascertain  the  hearing  from  station  1 
to  2  to  he  S.E.  85°  02'  37",  and  distance  =  350  feet ;  then,  crossing 
the  river  to  station  2,  we  run  and  measure  a  line  therefrom  to 
station  3,  situated  in  the  other  tangent  line ;  the  hearing  of  this 
line  we  determined  to  he  S.  72°  59'  3G"-41  E.,  and  its  length  = 
1604-264  feet;  then,  removing  to  station  3,  we  ascertain  the  hear 
ing  of  the  tangent  line  to  he  S.  60°  E.,  or  (which  is  the  same,) 
K.  60°  W. 

Having  thus  connected  the  straight  or  tangent  lines,  hy  a  traverse 
running  through  the  point  selected  as  the  most  suitable  for  the 
location  of  the  curve,  our  next  step  will  he  to  prepare  our  data  to 
ascertain  its  radius. 

Upon  examination  of  the  foregoing,  we  find  the  courses  and  dis 
tances  noted  in  the  following  table,  viz., 

BEARINGS.  DISTANCES. 

O         /         // 

Station  0  to  Station  1  =  S.  75  08  35-37  E.  =     152-5       feet 
"1  "          2  =  S.  85  02  37-00  E.  =     350-0         " 

"        2  "          3  =  S.  72  59  36-41  E.  =  l€04-264     " 

and  the  bearing  of  the  tangent  lines  from  0  to  A  due  East,  and  from  A  to  3  S.  60°  E. 

Having  thus  collected  and  arranged  our  courses  and  distances, 
we  then  compute  their  northings  and  southings,  eastings  and  west 
ings,  according  to  the  requirements  of  the  case. 


USEFUL      FORMULA. 

Eepresenting  the  northing  by  N,  southing  by  S,  easting  by  E, 
and  westing  by  W ;  and,  for  the  convenience  of  a  general  expression 
in  our  formula,  we  call  the  northings  or  southings  the  latitudes, 
which  we  represent  by  L  ;  and  the  eastings  and  westings  the  depar 
tures,  which  we  represent  by  D.  Then,  putting  E  for  the  radius 
of  the  tables,  B  for  the  bearings,  and  d  for  the  distance,  we  have 

II :  d  : :  sin.  B  :  D  =  sin.  B  .  d ; 
and  E  :  d  : :  cos.  B  :  L  =d.  cos.  B  (M) 

COMPUTATIONS     OF     LATITUDES     AND     DEPARTURES. 

No.  1.    B  =  S  750  08'  35" -37  E     sin.  =  9*9852331  cos.  =  9-4089262 

d  =       152-5  feet  log.  =  2  •  1832698  log.  =  2  •  1832698 


D  =       147-401  feet  log.  =  2-1685029        L  =     39-1017    log.  =  1-5921960 

No.  2.     B  =  S  85°  02'  37"  E  sin.  =  9-9983730  cos.  —  8-9365008 

d  =       350  feet  log.  =  2-5440680  log.  =  2-5440680 


D  =       348-691  feet  log.  =  2-5424410        L  =     30-239      log.  =  1-4805688 

No.  3.     B  =  S  720  59/  36//.41  E     sin.  =  9-9805811  cos.  =  9-4660977 

d  —        1604-264  feet          log.  =  3-2052779  log.  =  3-2052779 


D  =        1534-119   feet          log.  =  3-1858590        L  =  469-219      log.  =  2-6713756 

Having  computed  the  latitudes  and  departures,  or,  in  other 
words,  the  southings  and  eastings  indicated  by  the  tables  of  courses 
and  distances ;  we  then,  to  render  these  operations  as  perspicuous 
as  we  well  can,  re-arrange  in  a  tabular  form,  our  courses  and  dis 
tances,  with  the  southings  and  eastings  belonging  to  each  ;  and 
having  summed  them  up,  we  proceed  to  compute  the  bearing  and 
distance  from  0  to  station  3.  Thus,  by  making  use  of  the  symbols 
of  the  preceding  formula,  with  the  addition  of  5,  by  which  we  repre 
sent  the  distance  from  station  0  to  3,  we  have 

Tan.  B  =  -£   ;  and  sin.  B  :  D  : :  11  :  $  =  -J^-; 
or,  Cos.  B  :  L  : :  K :  a  =  J,; ,  (N) 


BEARINGS      OF      STATIONS. 


BEAEINGS. 

DISTANCES. 

SOUTHINGS. 

EASTINGS. 

Station  0  to  1 

S  75°  08'  35"-  07  E 

152-5      feet 

39-102 

147-401 

"       1  to  2 

S  85°  02'  37"-OOE 

350-0 

30-239 

348-691 

"       2  to  3 

S  72°  59'  36'/-  41  E 

1604-264   " 

469-219 

1534-119 

L  =  538-  560 

D  =  2030-  211 

D    =  sum  of  eastings  =  2030-211 

L     =  sum  of  southings        =     538-560 


log.  =  3-3075411 
log.   =  2-7312341 


Bearing  from  station  0  to  3  =  B  =  S  75°  08'  35"-37E          tan.  =  0-5763070 


B    =75°  08'  35" -37    sin.   =  9-9852331  cos.  =  9-4089261 

D    =  2030-211  log.   =  3-3075411        L  =  538-560        log.   =  2-7312341 


<5     =  2100-429  feet        log.  =  3-3223080        Proof 


log.  =  3-3223080 


Having  thus  obtained  the  "bearing  from  station  0  to  3,  viz., 
S  75°  08'  35" -37  E,  and  distance  =  2100-429  feet,  our  next  step 
will  he  to  ascertain  the  distances  0  A  and  A  3.  In  the  triangle 
A  3  0,  we  have  to  find  the  several  angles.  The  bearing  from 


o  to  A 
o  to  3 

A  to  o 
A  to  3 

3  to  A 
3  toO 


=    due  East 

=     S  75°  08'  35"- 37  E 


>  which  gives  angle  at  0    =    14°  51'  24". 


=  due  West  ) 

=  S  60°  00'  00"-OOE   ) 

=  N  60°  00' 00"- 00  W 

=  N  75°  03' 35". 37  W 


150°  00' 0  "-00 


15°  08' 35". 37 


Having  found  the  angles,  we  have 

Sin.  A  :  5  : :  sin.  0  :  A  3 
and  Sin.  A  :  8  : :  sin.  3  :  0  A 


S  .  sin.  0  . 
sin.  A  » 
.  sin.  3 


sin.  A 


Thus, 


114 


USEFUL      FORMULA. 


A  =    1500  00'  00" -00  co.  ar.  sin.  =  0- 

0    =    14°  51'  24? -63  sin.  =  9-4089261 

5    =  log.  =  3  •  3223080 

A  3  =  1077-12  feet  log.  =  3-0322641 


A      =                            co.  ar.  sin.  =  0-3010300 

6      =  log.  =  3/3223080 

3      =  150  08' 35" -37  sin.  =  9 -4170259 
AO  =  1097-4  feet 


log.  =  3-0403639 


For  the  purpose  of  ascertaining  the  angle  G  in  the  triangle 
A  G  C,  we  assume  a  radius  =  unity,  which  we  represent  in  our 
formula  by  1,  retaining  r  as  the  radius  in  the  unit  of  measure. 
Then,  putting  A  for  the  angle  at  apex  ;  T  and  T  for  the  tangent 
points  ;  C  for  the  angle  at  the  centre  of  the  curve ;  G  for  the  angle 
at  station  1  in  the  traverse ;  A'  for  the  angle  T'  A  G ;  C'  for  the 
angle  A  C  G ;  we  have  in  the  triangle  A  0  G,  the  angle  0  and  the 
sides  0  A  and  0  1,  to  find  the  side  A  1  =  A  G,  and  the  angles  A' 
and  G.  Putting  a  for  the  side  0  A,  and  b  for  the  side  0  1  =  0  G, 
we  have 

a  +  b  :  a  c~  b  ::   tan.  J  (180°  —  0)  :  tan.  J  (A'  co  G)  = 


(a  co  ft)  .  tan.  4-  (180°  —0) 


and  J  (180°  —  0)  +  J  (A'  ^  G)  =  G  ; 
and  \  (180°  —  0)  —  i  (A'  co  G)  =  A' 


(0) 


Having  found  the  angles  A'  and  G,  we  find  the  side  A  G,  which 
we  represent  by  m,  by  either  of  the  analogies  following : 

Sin.  A'  :  b  I :  sin.  0  :  m  =  —jj^r-  ; 
or,  Sin.  G  :  a  1 1  sin.  0  '.  m  =   —^n^0~  (P) 

Having  obtained  m,  and  putting  G'   for   the   angle   G  in    the 
triangle  A  C  G,  we  have 

Sin.  |  A:  1  ::E:AC=-sta1jX-; 
then,  representing  A  C  by  n,  we  have 

1  :  sin.  (|  A  —  A')  : :  n  :  sin.  G'  =  n  .  sin.  (J  A  —  A')        (Q) 


And  sin.  (J  A  —  A'  -f-  G')  :  m  : :  sin.  (J  A 

_  m  • sin-  (*  A  —  A/) 
sin.  (i  A  —  A'  +  G') 


A'):r  = 


(R) 


BEARINGS      OF  STATIONS.  115 

We'  now  introduce   an   example  of  computation   [formulae  (0) 
and  (P.)] 

a  =     1097 -4  feet  180°  00*  00"  -00 

b  =       152-5    "  0  =    14°51'24"-63 


a  +  b             =  1249-9    co.  ar.  log.   =  6-9031247                                          2)165°  OS'  35" -37 

a  ~>  b             =  944-9  log.   =  2-9753858             i  (180°  —  0)   =    82°  34'  17" -68 

i  (180°  —  0)  =  82°  34'  17" -68  tan.   =  0-8847831 

J  (A7  co   G)     =  80°   12'  52" -64  tan.   =  0-7632936 


G  =162°  47'   10" -32  G    =162°  47'   10" -32   co.  ar.   sin.   =  0-5287991 

A'  =2°  21'  25" -04  a    =  1097-4  feet  log.  =  3-0403650 

0  =     14°  51'  24" -63  0     =     14°  51'  24" -63  sin.  =  9-4089261 


Proof  =  180°  00'   00" -00  m    =     950-8   feet  log.   =  2*9780902 

By  formulae  (Q)  and  (R)  we  have 

i  A  =  75°  00'  00" -00  co.  ar.  sin.  =  0-0150562  =  n 

J  A  —  A'  =72°  38'  34" -96  sin.   =  9-9797599 

Ambiguous  G'  =  81°  09'  53" -75  *  sin.   =  9-9948161 

True  G'  =98°  50'  06" -25 

i  A  —  A'  =  72°  38'  34". 96 


i  A  —  A  +  G'  =  171°  28'  41" -21  co.  ar.  sin.   =  0-8291893 

m  =950-8  feet  log.   =  2-9780892 

k  A  —  A'          =  72°  38'  34" -96  sin.   =  9-9797599 


r  =6124-05  feet  log.  =  3-7870384 

Having  ascertained  the  radius  which  the  problem  requires,  we 
proceed  to  ascertain  the  deflection  for  a  chord  of  50  feet. 

By  formula  (3)  we  have  sin.  D  =  _*£*.  •  hence 

r  =  6124-05  feet  co.   ar.  log.  =  6-2129616 

4  ch     =       25  "  log.  =    1-3979400 

I)  =  0°  14'  02"  sin.   =  7-6109016 

*  As  the  problem  under  all  its  forms  requires  G'  to  be  larger  than  a  right  angle,  it  is  evident  that 
the  true  G'  must  be  the  supplemental  angle,  inasmuch  as  the  sine  of  an  angle  is  the  same  as  the  sine 
of  its  supplement. 


116  USEFUL      FORMULAE. 

The  deflection  thus  found  being  an  awkward  sum  to  add  or  sub 
tract  in  the  field,  we  may  assume  one  more  convenient  without 
materially  changing  the  location  of  the  curve ;  we  therefore  assume 
0°  14'  as  the  measure  of  a  deflection ;  then,  by  formula  (5)  we 
have  r  =  — \^  ;  and,  for  the  purpose  of  ascertaining  by  what 
amount  this  change  in  the  length  of  the  radius  will  affect  the  loca 
tion  of  the  curve,  we  will  endeavor  to  find  the  distance  from  the 
apex  to  the  middle  of  the  curve  for  each  radius.  By  formula  (6) 
we  have  t  =  tan.  \  C  .  r ;  and  by  (7)  we  have 
b  ==  t.  tan.  J  C  ==  tan.  J  C  .  tan.  J  C  .  r  =  '»»•  * c  .J 


THE    VALUE    OP    *,  COMPUTED    FROM    THE    RADIUS 


ALREADY    OBTAINED. 


\  C    =  15°  00'  00"  tan.  =  9-4280525 

i  C    =     7°  30'  00"  tan.  =  9-1194291 

r          =6124-05  log.   =  3-7870384 


b         =     216-033  feet  log.  ==  2-3345200 


THE  VALUE  OF  b,  COMPUTED  TO  CORRESPOND  TO 

A  RADIUS  BASED  UPON  A  DEFLECTION  OF  0°   14'. 

D       =  00  14'  00"  co.  ar.  sin.  =  2-3901470 

J  C    =  15°  007  00"  tan.  =  9-4280525 

i  C    =  70  30'  00"  tan.  =  9-1194291 

4-  ch  =  25  feet  log.  =  1-3979400 


=  216-555  feet  log.   =  2-3355686 

216-033      " 


0-522      " 

Thus  we  see  that  the  proposed  change  in  the  deflection  will  affect 
the  location  of  the  curve  only  0*522  feet,  an  amount  in  most  cases 
too  small  to  produce  any  practical  inconvenience. 

Having  shown  that  whenever  convenience  requires  a  change  of  a 
few  seconds  in  the  angle  of  deflection,  the  change  may  be  made 
without  materially  affecting  the  location  of  the  curve,  we  now  pro 
ceed  to  determine  a  radius  which  shall  correspond  with  the  desired 
deflection,  viz.,  of  0°  14',  as  explained  in  the  foregoing.  By 
formula  (5)  we  have  r  =  -f^-  Thus, 

D         =  0°  14'  co.  ar.    sin.   =  2-3901470 

i  ch.     =      25  feet  log.   =   1-3979400 

r  =  6138-853  feet  log.   =  3-7880870 


BEARINGS      OF      STATIONS.  117 

To  find  the  position  of  the  tangent  points  at  stations  T  and  T 
we  compute  their  distance  from  apex,  and  compare  them  with  the 
distances  of  0  and  3,  which  have  been  already  determined.  By 
formula  (G)  we  have  t  =  tan.  J  C  .  r.  Thus, 

r  =  6138-853  feet  log.  =  3-7880870 

J  C      =  15°  00'  00"  tan.  =  9-4280525 


t  =  1644-90  log.   =  3-2161395 

A  to  3,  heretofore  computed  —  1077-12 
567-78 

We  thus  find  the  point  T  567 '78  feet  further  from  A  than  the 
point  3.     Again, 

t  =     1644-90  feet 

A  to  0,  heretofore  computed   =     1097-40    " 


547-50    " 


We  thus  find  the  point  T  547-50  feet  further  from  A  than  the 
point  0. 

To  find  the  length  of  arc  from  point  T  to  point  G,  corresponding 
to  point  1  in  the  traverse,  we  have  in  the  triangle  A  C  G,  the  angle 
at  C  =  180°  —  (the  angle  G'  +  angle  A.)  Thus,  we  find 

A   =  (£A  —  A')  =  75°  —  2°  21'  25"'04   =  72°  38'  34"'96 
G'  =980  50'  06"  -25 

C  =  the  supplement  =    8°  31'  18"  -79 


Proof  180°  00'  00" -00 

Then,  we  have  half  the  centre  angle  =  \  C  =  15°  00'  00"  minus 
the  supplementary  angle  C  found  above  =  the  angle  C  in  the 
triangle  GCT.  Thus, 

*  C  =15°   00'  00" -00 

Supplementary  C  =    8°  31'  18" -79 


Hence  the  C  sought  6°  28'  41" '21 


BEARINGS      OF      STATIONS.  119 

Then,  representing  the  angle  C  thus  found  and  reduced  to  seconds 
by  C",  and  the  arc  sought  by  a,  we  have,  by  formula  (8,)  a  =  -—£— 

C"  =  23321-21  log.  =  4-3677511 

r  =  log.   =  3-7880870 

r"  =  co.  ar.    log.  =  4 '6855749 

a  =      694-09  feet             log.  =  2'8414130 

If  we  now  consider  T'  as  numbered  3*06,  in  the  stations  of  loca 
tion,  and  the  numbers  in  the  location  to  be  increasing  as  we  enter 
the  curve,  we  shall  find  the  point  at  G,  or  rather  near  G,  (as  we 
have  slightly  changed  the  radius,)  to  be  equal  to  3 -06  -f-  6 '94, 
which  increases  the  number  of  the  locating  stations  to  10.  For  the 
purpose  of  avoiding  fractions,  we  ascertain  the  point  T'  by  measure 
ment  from  0,  and  then  locate  the  curve  to  station  10 ;  from  this 
point  we  ascertain  the  direction  of  the  radius,  and  select  the  point 
A,  which  should  be  so  situated  as  to  command  a  distinct  view  of 
the  locality  where  the  work  is  to  be  laid  out.  We  then  discover  the 
relative  direction  and  length  of  the  line  10  A,  by  ascertaining  the 
angle  which  it  makes  with  the  radius  of  the  curve,  and  measuring 
the  distance  between  the  termini.  Let  us  now  suppose  the  angle 
to  measure  20°  00',  and  the  length  of  the  line  to  be  200  feet.  We 
next  cross  the  river  or  bay,  and  select  the  point  B,  which  should 
likewise  command  a  distinct  view  of  the  locality  where  the  work  is 
to  be  laid  out ;  we  then  measure  the  angles  which  form  the  triangle 
A  B  10,  and  compute  their  relative  positions.  Now,  supposing  the 
angles  at 


830  00' 
32°  0(X 

650  0(X 
1800  00' 


120  USEFUL      FORMULA. 

Then,  putting  5  for  the  distance  between  10  and  A,  we  have 
Sin.  B  :  6  : :  sin.  10  :  A  B  ;         and  sin.  B  :  6  : :  sin.  A  :  10  B     (S) 
Thus, 

B  =32°  00'  co.  ar.    sin.  =  0-2757903 

S  =  200  feet  log.  =  2-3010300 

10  =65°  0(X  sin.  =  9-9572757 

A  B  =  342-05  feet  log.  =  2-5340960 

j 

log.  =  2-5768203 


Sin.  B 

A  =  830  00'  sin.   =  9-9967507 


10  B        =  374-61  feet 

(50)  We  have,  from  station  10  to  10  •  50,  a  distance  of  50  feet 
of  arc;  then,  from  10 '50  to  11,  a  like  distance;  and  so  on,  from 
station  to  station,  to  station  13  •  50.  (See  Fig.  20.)  To  compute  in  the 
readiest  manner  the  relative  position  of  these  several  points,  or 
rather  the  relative  positions  of  the  points  a  b  c  and  c?,  (which  repre 
sent  the  corners  of  the  piers  situated  about  these  stations,)  we  assume 
the  radius  of  the  curve  from  station  10  to  bear  due  south,  and  C 
as  a  zero  point.  Then,  by  ascertaining  the  distances  and  relative 
bearings  to  each  of  the  points,  we  compute  what  we  shall  (for  the 
want  of  more  appropriate  terms)  call  their  northings  or  southings, 
eastings  or  westings,  without  consideration  of  their  astronomical  or 
geodetical  bearings.  To  ascertain  the  angles  of  the  radii  from  C  to 
these  several  points,  we  put  a  for  the  arc  connecting  them,  r  for  the 
radius  in  the  unit  of  measure,  and  r"  for  an  arc  in  seconds  equal  in 
length  to  radius.  We  then  have,  by  formula  (9,)  C"  =  —^- 
Thus  the  angle  at  C,  between  stations  10  and  10-50,  gives  a  =  50 
feet.  Then, 

a       =50  feet  log.  =  1-6989700 

r"      =  log.   =  5-3144251 

r       =  co.  ar.    log.  =  6-2119130 


C"    =  1680"  =  nearly  log.  =  3-2253081 


BEARINGS      OF      STATIONS.  121 

and  1680",  reduced  to  degrees  and  minutes,  will  give  =  0°  28'  00". 
We  have  given  this  computation  for  the  purpose  of  explaining  a 
general  rule  which  will  apply  in  all  cases. 

In  the  present  case,  the  arc  a  =  the  chord  of  one  of  our 
deflections ;  and,  as  the  difference  between  the  chord  of  fifty  feet 
in  length  and  the  arc  it  spans  (based  upon  a  radius  of  6138*853 
feet)  is  so  small  that  the  one  may  be  taken  for  the  other,  in  the 
practical  operations  of  locating  a  railroad,  we  therefore  may,  without 
sensible  error,  take  the  angle  at  C  for  50  feet  of  arc  =  two 
deflections  =  28',  or  the  same  as  above. 

But  we  shall,  notwithstanding,  when  we  come  to  consider  .the 
dimensions  of  the  piers,  find  a  necessity  for  the  formulae.  Let  us 
assume  the  foundations  of  the  piers  to  be  8  feet  broad  and  18  feet 
in  length.  Now,  as  the  stations  named  above  correspond  to  the 
centre  of  these  piers,  we  find  it  necessary  to  determine  half  the 
angular  width  of  them  from  C.  Thus, 

a  =4  feet  log.  =  0*6020600 

r"  =  log.  =  5-3144251 

r  co.  ar.    log.  =  6-2119130 

•      C"  =   134"- 4  log.  =  2-1283981 

Reducing  C"  to  degrees  and  minutes,  gives  us  the  angle  = 
0°  2'  14"  '4: ;  but,  for  the  purpose  of  avoiding  (in  the  computations) 
the  fractions  of  a  second,  we  may,  without  varying  the  dimensions 
of  the  piers  perceptibly,  assume  the  angular  width  of  the  half  pier 
to  be  0°  2'  15".  For  like  reasons,  with  a  radius  of  the  length  we 
have  adopted,  we  assume  both  ends  of  the  pier  to  be  of  the  same 
angular  width. 


122  USEFUL      FORMULA. 

Having  made  these  explanations,  we  proceed  to  construct  a  table 
of  the  angular  positions  of  the  corners  of  the  piers  represented  by 
abed.  We  have  stated  above  that  the  centres  of  the  piers  are  two 
deflections,  or  28',  apart  from  C.  Then,  taking  station  10  for  a 
starting  point,  and  the  radius  from  point  C  through  this  point,  as 
bearing  due  north,  we  have  the  angle  to  station  10 -50  =  28'  ;  and 
the  angle  to  station  11,  twice  28'  ;  and  so  on.  Having  determined 
the  position  of  the  primitive  stations,  we  may,  by  additions  and 
subtractions  of  the  angular  half  widths  and  .widths  of  the  piers, 
determine  the  angular  positions  of  the  points  abed;  and  upon 
these  principles  we  construct  the  following  table,  viz., 

RELATIVE                          RELATIVE  RELATIVE 

BEARINGS  OF  BEARINGS  OP  BEARINGS  OF 

PRIMITIVE  STATION                   b  AND  d  a  AND  c 

N.E.                                  N.E.  N.E. 

Station  10  to  10-50  angle  =     28'  -f  2'  15"  =        30'  15" 

"         "  11-00  «  =     56'+       "       =         58'  15'/ —  4' 30"  =        53' 45" 

"  11-50  "  =     84'+       "       =  1°26'15"  —  "  =  1°21'45" 

"        "  12-00  "  =  112'  +       "       =  1°  54'  15" —  "  =  1°  49'  45" 

"        "  12-50  "  =140'+       "       =  2°  22'  15" —  "  =  2°  17' 45" 

«         «  13-00  "  =168'+       "       =  "  *=2°45'45" 

«  13-50  "  =  196'  =  N.E.  3°  16' 


Having  thus  arranged  a  table  of  bearings  from  the  centre  of  the 
curve,  or  C,  of  a  b  c  d,  with  the  primitive  station  to  which  they  are 
connected,  we  next  prepare  a  table  containing  both  bearings  and 
distances,  leaving  a  space  for  the  northings  and  eastings  to  be  added 
after  computation. 


BEARINGS      OF      STATIONS. 


123 


A     TABLE     OF     BEARINGS, 

DISTANCES,     NORTHINGS    AND     EASTINGS,     OF     STATIONS, 
FROM     C,     OR     THE     CENTRE     OF     THE     CURVE. 


PRIMITIVE  STATIONS. 

BEARINGS. 

DISTANCES. 

NORTHINGS 

IN   FEET. 

EASTINGS 

IN   FEET. 

10  -00  from  C  to  10 

Due 

North 

r  +  0  feet 

=  6138-85  feet 

6138-85 

00-000 

10-50    "      "  "  d 

N.E. 

=  00  30'  15" 

r  —  9    " 

=  6129-85     " 

6129-61 

53-938 

"       «      "  "  b 

a 

«         « 

r  +  9    « 

=  6147-85    " 

6147-61 

54-096 

11-00    "      "  "   c 

N.E. 

=  QO  53'  45" 

r  —  9    " 

=  6129-85    " 

6129-10 

95-838 

"       «      «  "  a 

u 

(C              it 

r  +  9    " 

=  6147-85    " 

6147-10 

96-119 

U             i(          t(    (C     d 

N.E. 

=  QO  58'  45" 

r  —  9     " 

=  6129-85    " 

6128-97 

103-860 

"       "      "  "  b 

a 

u          u 

r  +  9    " 

=  6147-85    " 

6146-97 

104-165 

11-50    "      "  "  c 

N.E. 

=  10  21'  45" 

r  —  9    " 

=  6129-85     " 

6128-12 

145-755 

d 

" 

it             U 

r  +  9    " 

=  6147-85    " 

6146-11 

146-183 

"    "   "  "  d 

N.E. 

=  10  26'  45" 

r  —  9    " 

=  6129-85     " 

6127-92 

153-176 

"       "      "  "  b 

u 

u        a 

r  +  9    " 

=  6147-85    " 

6145-92 

154-228 

12-00    "      "  "   c 

N.E. 

=  10  49'  45" 

r  —  9    " 

=  6129-85     " 

6126-73 

195-662 

"       "      "  "  a 

a 

a          u 

r  +  9    " 

=  6147-85    " 

6144-72 

196-237 

"       "      "  "  d 

N.E. 

=   10  54'  15" 

r  —  9    " 

=  6129-85     " 

6126-46 

203-682 

"       "      "  "  b 

a 

u          u 

r  +  9    " 

=  6147-85     " 

6144-45 

204-280 

12-50    "      "  "  c 

N.E. 

=  2°  17'  45"         r  —  9    " 

=  6129-85     " 

6124-93 

245-557 

"       "      "  "  a 

ic 

a          a 

r  +  9    " 

=  6147-85     " 

6142-92 

246-177 

"       "      "  "  d 

N.E. 

—  20  22'  15" 

r  —  9    " 

=  6129-85     " 

6124-60 

253-574 

"       "      "  "  b 

u 

u          u 

r  +  9    " 

=  6147-85    " 

6142-59 

254-318 

13-00    "      "  "  c 

N.E. 

=  2  045'  45" 

r  —  9    " 

=  6129-85     " 

6122-81 

295-435 

"       "      "  "  a 

u 

11             U 

r  +  9     « 

=  6147-85    " 

6140-79 

296-302 

13-50    "      "  "   0 

N.E. 

=  30  16'  00" 

r  —  0     " 

=  6138-85     " 

6128-87 

349-811 

Point  A  from  C 

5950-91 

—  68.404 

"     B     "      " 

• 

5873-97 

269-884 

Having  prepared  our  table  of  bearings  and  distances  from  the 
centre  C  to  the  corners  of  the  piers,  we  introduce  examples  of 
computation  of  northings  and  eastings  ;  all  the  bearings  the  table 
contains  being  north-east.  Substituting  <5  for  d,  we  have,  by 
formula  (M,) 

D  =  sin.  B  <5 ;  and  L  =  cos.  B  «5. 


124  USEFUL      FORMULAE. 

10-50    d    =  N.E.    0°  30'  15"    sin.  =  7'9444459  cos.  =  9'9999832 

(5    =  6129-85         log.  =  3-7874499  log.  =  3-7874499 


D  =  53-938  log.  =  1-7318958        L  =  6129-61        log.  =  3-7874331 

10-50    b    =  N.E.  sin.  =  7-9444459  cos,  =  9'9999832 

d    =  6147-85         log.  =  3-7887233  log.  =  3-7887233 


D  =  54-096  log.  =  1-7331692        L  =  6147-61        log.  =  3-7887065 

11-00    c    —  N.E.    00  53'  45"    sin.  =  8-1940869  cos.  =  9'9999469 

(5    =  6129-85         log    =  3-7874499  log.  =  3-7874499 


D  =  95-838  log.  =  1-9815368        L  =  6129-10        log.  =  3  "7873968 

11-00    a    =  N.E.    0°  53' 45"    sin.  =  8-1940869  cos.  =  9-9999469 

6    =  6147-85         log.  =  3-7887233  log.  =  3-7887233 


D  =  96-119  log.  =  1-9828102        L  =  6147 '10        log.  =  3-7886702 

11-00    d    =  N.E.    Q°t58f  15"    sin.  =  8-2290013  cos.  =  9"9999377 

<J    =  6129-85         log.  =  3-7874499  log.  =  3-7874499 


D   =  103-86  log.  =  2-0164512        L  =  6128-97        log.  =  3'7873876 

11-00    b    =  N.E.    00  58'  15"    sin.  =  8-2290013  cos.  =  9-9999377 

(5    =  6147-85         log.  =  3-7887233  log.  =  3-7887233 


D  =  104-165         log.  =  2-0177246        L  =  6146-97        log.  =  3-7886610 

11-50    c    =  N.E.    10  21'  45"    sin.  =  8-3761729  cos.  =  9.9998772 

d    =  6129-85         log.  =  3-7874499  log.  =  3-7874499 


D  =  145-755        [log.  =  2-1636228        L  =  6128-12        log.  =  3'7873271 

11-50    a    =  N.E.    10  21'  45"    sin.  =  8-3761729  cos.  =  9*9998772 

d    =  6147-85         log.  =  3-7887233  log.  =  3-7887233 


D  =  146-183         log.  =  2-1648962        L  =  6146-11        log.  =  3-7886005 

11-50    d    =  N.E.    1°  26'  15"    sin.  =  8-3994397  cos.  =  9-9998633 

«5    =  6129-85         log.  =  3-7874499  log.  =  3-7874499 


D=  153-776         log.  =  2-1868896        L  =  6127*92        log.  =  3-7873132 


BEARINGS      OF      STATIONS.  125 


11-50  b  =  N.E.  10  26'  15"  sin.  =  8 '3994397  cos.  = 

J  =      6147-85    log.  =  3-7887233  log.  =  3-7887233 


D=      154-288    log.  =  2-1881630   L  =  6145-92   log.  =  3-7885866 

12-00  c  =  N.E.  1°  49'  45"  sin.  =  8*5040569  cos.  =  9-9997786 

d  =      6129-85    log.  =  3-7874499  log.  =  3-7874499 


D  =  195-662  log.  =  2-2915068  L  =  6126"73   log.  =  3'7872285 

12-00  a  =  N.E.  1°  41'  45"  sin.  =  8-5040569  cos.  =  9-9997786 

r5  =  6147-85  log.  =  3-7887233  log.  =  3-7887233 

D  =  196-237  log.  =  2-2927802  L  =  6144-72   log.  =  3-7885019 

12-00  d  =  N.E.  10  54'  15"  sin.  =  8'5215024  cos.  =  9'9997601 

(J  =  6129-85  log.  =  3-7874499  log.  —  3-7874499 


D  =      203-682    log.  =  2 "3089523   L  =  6126-46   log.  =  3 "7872100 

12-00  b  =  N.E.  1°  54'  15"  sin.  =  8-5215024  cos.  =  9-9997601 

(5  =      6147-85    log.  =  3-7887233  log.  =  3-7887233 


D  =      204-280    log.  =  2-3102257   L  =  6144*45   log.  =  3- 

12-50  c  =  N.E.  20  17' 45"  sin.  =  8-6027015  cos.  =  9-9996513 

(5  —      6129-85    log.  =  3-7874499  log.  =  3 -7874499 


D  =      245-557    log.  =  2'3901514   L  =  6124-93   log.  =  3-7871012 

12-50  a  =  N.E.  2 °  17'  45"  sin.  =  8-6027015  cos.  =  9-9996513 

d  =s      6147-85    log.  =  3-7887233  log.  =  3-7887233 


D  =      246-177    log.  =  2-39142-18   L  =r  6142' 92   log.  =  3-7883746 

12-50  d  —  N.E.  20  22'  15"  sin.  —  S'6166545  cos.  =  9*9996281 

d  =      6129-85    log.  =  3-7874499  log.  =  3-7874499 


D  =      253-574    log.  =  2-4041044   L  =  6124'60   log.  =  3-7870780 

12-50  b  =  N.E.  20  22'  15"  sin,  =  8'6166545  cos.  =  9-9996281 

f5  =      6147-85    log.  =  3-7887233  log.  —  3  '7887233 


D=      254-318    log.  =  2-4053778   L  =  6142-59   log.  =  3-7883514 


126 


USEFUL      FORMULA. 


13-00  c  ==  N.E.  20  45'  45"  sin.  =r  8'6830114 
d  =  6129-85  log.  =  3-7874499 
D  =  295-435  log.  =  2*4704613  L  =  6122-81  log.  =  3-7869510 


cos.  =  9-9995011 
log.  =  3-7874499 


cos.  =  9-9995011 
log.  =  3-7887233 


13-00    a    =  N.E.    2°  45'  45"    sin.  =  8-6830114 
d    =  6147-85         log.  =  3-7887233 

^ 

D=  296-302         log.  =  2-4717347        L  =  6140-79        log.  =  3-7882244 

13-50  station  N.E.    3°  16'  00"    sin.  =  8-7557469  cos.  =  9-9992938 

d    =  6138-85         log.  =  3-7880870  log.  =  3-7880870 

D  =  349-811         log.  =;  2-5438339        L  —  6128-87        log.  =  3-7873808 

Having  computed  all  the  points  connected  with  the  centre  C,  and 
carried  the  results  into  the  preceding  table  of  bearings,  distances, 
etc.,  we  next  compute  the  relative  situation  of  the  points  A  and  B 
from  station  10.  We  have  before  stated  that  the  line  10  A  made 
an  angle  with  the  radius  of  the  curve  of  20° ;  the  radius  being  taken 
to  bear  due  south  from  10,  gives  the  bearing  of  10  A  =  S.  20°  W. 
The  angle  at  A,  in  the  triangle  10  A  B,  being  83°,  gives  the  bearing 
of  A  B  =  S.  77  E.  And  the  angle  at  10,  being  65°,  gives  the 
bearing  10  to  B  =  S.  45°  E.  Having  thus  ascertained  the  bearings, 
and  the  distances  being  already  computed,  we  now  make  up  a  table 
of  bearings  and  distances,  leaving  room  to  put  in  the  latitudes  and 
departures  when  obtained. 

BEARINGS    AND     DISTANCES, 
FROM     STATION     10    TO     A    AND     B,    AND     FROM    A     TO    B. 


BEARINGS  AND  DISTANCES. 

SOUTHINGS. 

EASTINGS. 

WESTING. 

From    10  to  B  =    S.E.    45°    =  374-61  feet 

264-884 

264-884 

"        10  to  A  =    S.W.  20°    =  200-00  feet 

187-939 

68-404 

"        A   to  B  =    S.E.     770    =  333-288  feet 

76-945 

338-288 

We  now  proceed  to   compute  the  latitudes  and  departures. 


BEARINGS      OF      STATIONS.  127 

COMPUTATIONS    OF     THE     LATITUDES    AND     DEPARTURES. 

S.E.  450  00'  00"   sin.  =  9-8494850  cos.  =  9-8494850 

d  =  374-61     log.  =  2-5735710  log.  =  2-5735710 


D  =  264-884     log.  s=  2-4230560   L  =  264-884   log.  =  2-4230560 

S.W.  20°  00'  00"  sin.  =  9-5340517  cos.  —  9-9729858 

d  =  200  feet    log.  =  2-3010300  log.  =  2-3010300 


D  =  68-404     log.  =1-8350817   L—  187-939   log.  =2-2740158 

S.E.   77°  00'  00"  sin.  =  9 '9887239  cos.  =  9-3520880 

d  =  342-05     log.  =  2-5340960  log.  =  2-5340960 


D  =  333-288     log.  =  2-5228199   L  =  76'945    log.  =  1-8861840 

Having  computed  our  latitudes  and  departures,  and  written  them 
in  the  above  table,  it  now  becomes  necessary  to  ascertain  their 
relative  position  to  the  point  C.  We  find  by  our  table,  computed 
from  the  point  C,  that  station  10  has  a  northing  of  6138 '853  feet, 
and  easting  it  has  nothing.  We  see  by  the  above  table  that  A  is 
west  of  10  =  G8'404  feet.  For  the  purpose  of  preventing  the 
necessity  of  an  additional  column,  we  shall  note  this  in  our  table 
of  northings  and  eastings  from  point  C,  as  minus  eastings,  distin 
guishing  it  by  the  negative  sign,  thus,  ( —  68*404,)  in  the  easting 
column.  Now  A,  being  south  of  station  10  187*939  feet,  we 
subtract  this  sum  from  the  northing  of  10,  which  gives  the  northing 
of  A.  Thus, 

Station  10     northing  =        6138-853 

A    from        10  "         =    —  187-939 


Leaves  the  northing  of  A   =        5950-914  feet 


Then,  managing  in  a  like  manner  with  station  B,  we  have  B  east 
of  10  =  264-884  feet ;  and,  as  10  has  neither  easting  nor  westing, 
we  put  this  into  our  table  as  the  proper  easting. 


128  USEFUL      FORMULA. 

For  the  northing  we  have 


Station          10  =        6138-853 

and       "  B  =     —  264-884 


Leaves  the  northing  of  B  =        5873-969 

Again:  for  the  purpose  of  proving  a  portion  of  our  work,  we 
ascertain  the  relative  position  of  B  from  A.  We  have  previously 
found  the  situation  of  A  to  be 

A  northing    =        5950-914     and  easting  —    68-404 

B  (from  the  preceding  table)      "  =    —    76-945  "  338-288 


B  «  =        5873-969  "  269 '884 

Adding,  as  indicated  by  the  algebraic  signs,  gives  the  northings 
and  eastings  of  B  as  above.  We  now  carry  these  results  into  the 
preceding  table  of  northings  and  eastings,  etc.,  from  C ;  and  it  will 
then  contain  all  the  points  which  are  needed. 

Having  completed  our  table  of  bearings,  distances,  northings,  and 
eastings,  etc.,  from  the  point  or  centre  C,  we  next  compute  the 
bearings  from  A  and  B  to  the  several  corners  of  each  pier  and 
abutment  noted  in  the  aforesaid  table,  and  to  station  13 -50  of  the 
general  location ;  the  bearing  to  station  10  being  already  known. 
Beginning  with  corner  b,  in  the  abutment  at  station  10 '50,  the 
formula  may  be  thus  enunciated,  applying  the  affix  N  to  the 
expression  representing  the  station,  for  northing,  and  E  for  casting, 
we  then  have 

AN  co  £N  :  1  : :  AE  co  JE  :  tan.  B  =  -fj^-ff- 
wherein  B  expresses  the  bearing  sought. 


BEARINGS      OF      STATIONS. 


129 


EXAMPLE      OF     COMPUTATIONS. 

Station  10-50    AN  «~  &N  =      5950-914  co  6147-61    =  196*696  log. 

AE  co  bE  =    —  68-404  co      54-096  =  122 '500  log. 

B'  or  bearing  from  A  to  10 -50  b    =  N.E.  31°  54'  51"  *  tan.  =:  9-7943405 


2-2937956 
2-0881361 


Station   10-50    Ax  <~  <L\  =      5950'914  co  6129-61    =  178-696  log. 

AE  co  dE  =    —  68-404  co      53-938  =  122-312  log. 

B'  or  bearing  from  A  to  10'50  d    =  N.E.  31°  23'  49"      tan. 


Station  11-00  AN  cs>  ON  =  5950-914  co  6147-10  =  196-186 
AE  co  aR  —  —  68-404  co  96-119  =  164-523 
Bearing  from  A  to  11-00  a  =  N.E.  39°  59'  00" 

Station  11-00  AN  co  &N  =  5950-914  ^  6146-97  =  196-056 
AE  co  bE  =  —  68-404  co  104-165  =  172-569 
Bearing  from  A  to  11 -00 


:  2- 2521148 
;  2-0875756 
9-8354608 

log.  =  2-2S2G680 
log.  =  2-21C22G6 
tan.  —  9 -£235586 

log.  =  2-2923301 
log.  =  2-2369628 
=  N.E.  41°  21'  15"    tan.  =  9-9445827 


Station  ll'OO  AN  co  CN  =  5950-914  co  6129-100  =  178-186 
AE  co  CE  =  —  68-404  co  95'838  =  164'242 
Bearing  from  A  to  11-00  c  =  N.E.  42°  40'  29" 

Station  11-00  AN  co  dx  =  5950'914  co  6128-97  =  178-056 
AE  co  dE  =  —  68-404  co  103-860  =  172*264 


log. 
log. 


2-2508736 
2-2155842 
tan.  =  9-90471G6 


log. 
log. 


2-2505566 
2-2361945 


Bearing  from  A  to  11-00  d  =  N.E.  44°  03'  10"  tan.  •=  9-9856379 

Station  11-50    AN  co  «x  •==      5950-914  co  6 146 -11    =  195-196  log.  =  2*2904709 

AE  co  flE  =    —  68-404  co  146-183  =  214-587  log.  =  2-3342064 

Bearing  from  A  to  11-50  a  =  N.E.  47°  52'  48"  tan.  =  0-0437355 

Station  11-50    AN  co  &N  =      5950-914  co  6145-920  =  195-006  log.  =  2-29C0480 

AE  co  bs  =    —  68-404  co  154-228  =  222 '632  log.  =  2-3475876 

Bearing  from  A  to  11-50  b  =  N.E.  48°  47'  04"  tan.  =  0-0575396 


*  The  northing  of  A  being  less  than  the  northing  of  6,  and  the  easting  of  A  being  less  than 
the  bearing  must  of  course  be  northeasterly. 


130  USEFUL      FORMULAE. 

Station  11-50    Ax  <*>  ex  =      5950'914  co  6123' 120  =  177-206  log.  =  2-2184784 

AE  co  CE  =  —    G8'404  co     145-755  =  214-159  log.  =  2-33G7364 


Bearing  from  A  to  11-50  c          =  N.E.  50°  23'  38"          tan.  =  0-ce22580 

Station  11-50    Ax  co  dx  =      5950-914  co  G127'920  =  177-006  log.  =  2-2179880 

AE  co  dE  =    —  68-404  co     153-176  =  221-580  log.  =  2  -3455306 


Bearing  from  A  to  11-50  d          =  N.E.  51°  22'  56"  tan.  =  0-0975426 

Station  12-00    Ax  co  ax  =      5950-914  co  6144-720  =  193-806  log.  =  2-2873672 

AE  co  CE  —    —  68-404  co     196-237  =  264-641  log.  =  2-4226571 

Bearing  from  A  to  12-00  a          —  N.E.  53°  47'  00"  tan.  =  0-1352899 

Station  12-00    Ax  co  fcx  =      5950-914  co  6144-450  =  193-536  log.  =  2-2867617 

AE  co  IE  =    —  68-404  co     204'2SO  =  272 '684  log.  =  2 '4356597 


Bearing  from  A  to  12-00  b          =  N.E.  54°  38'  06"         tan.  =  0' 1488980 

Station  12-00    Ax  co  ex  =      5950-914  co  6126-730  =  175-816  log.  =  2-2450584 

AE  co  CE  =    —  68-404  co     195-662  =  264-066  log.  =  2-4217125 


Bearing  from  A  to  12-00  c          =  N.E.  56°  2(X  39"         tan.  =  0-1766541 

Station  12-00    Ax  co  ds  =      5950-914  co  6125-460  =  175-546  log.  =  2-2447125 

AE  co  dE  =    —  68-404  ^    203-682  =  272-086  log.  =  2-4347062 


Bearing  from  A  to  12-00  d          =  N.E.  57°  09'  04"         tan.  =  0-1899937 

Station  12-50    Ax  co  flx  =      5950-914  co  6142-920  =  192-006  log.  =  2-2833148 

AE  co  OE  =    —  68-404  co    246-177  =  314-581  log.  =  2-4977325 


Bearing  from  A  to  12'50  a          =  N.E.  58°  35'  37"        tan.  =  0-2144177 

Station  12-50    Ax  co  b*  =      5950-914  co  6142-590  =  191-676  log.  =  2-2825677 

AE  co  &E  =    —  68-404  co    254-318  =  322 '722  log.  =  2-5088286 


Bearing  from  A  to  12-50  ft          =  N.E.  59°  17'  33"        tan.  =  0-22G2609 

Station  12-50    Ax  co  Cx  =      5950-914  co  6124  930  ra  174-016  log.  =  2-2105892 

AE  co  CE  =    —  68-404  co    215-577  =s  313-961  log.  =  2-4968757 


Bearing  from  A  to  12-50  c          =  N.E.  61°  00'  08"        tan.  =  0-2562865 


BEARINGS      OF      STATIONS. 


131 


Station  12-50    Ax 

AE 


dx  =      5950-914  ^  6124-600  =  173-686 
dE  =    —  68-404  <^>    253-574  =  325  '978 


log.  =  2-2397648 
log.  =  2-5078262 


Bearing  from  A  to  12-50  d          =  N.E.  61°  39'  22"        tan.  =  0-2680614 


Station  13-00    Ax  <*>  ax  =      5950-914 
AE  co  aE  =    —  68-404 


6140-790  =  189-876 
296-302  =  364-706 


log.  =  2-2784701 
log.  =  2-5619430 


Bearing  from  A  to  13-00  a          =  N.E.  62°  29'  50"        tan.  =  0--834729 

Station  13-00    Ax  co  ex  =      5950-914  co  6122-810  =  171-896  log.  =  2-2352658 

AE  co  CE  =    —  68-404  co    295-435  =  363-639  log.  =  2-5609065 

Bearing  from  A  to  13-00  c          =  N.E.  64°  42'  41"        tan.  =  0'3256407 

Station  13-50    Ax  co  13'50x  =      5950-914  co  6128-870  —  177-956  log.  =  2-2502980 

AE  co  13-50K  =    —  68-404  co    349-821  =  418-215  log.  =  2-6213996 

Bearing  from  A  to  station  13-50  =  N.E.  66°  57'  01"  tan.  =  0-3711016 

Having  computed  the  bearings  from  A  to  the  corners  of  each  pier 
and  abutment  shown  in  the  diagram,  including  stations  10  and  13-50 
in  the  alinement  of  the  road,  we  now  proceed  to  compute  the  bearing 
of  the  corners  of  each  pier,  and  the  stations  10  and  13 -50  from 
station  B. 


Station  10-50    Bx  co  bs    =    5873-969  co  6147-610  =  273-641 
BE  co  frE    =      264-881  co  .  54-096  =  210-783 


log.  —  2-4371812 
log.  =  2-3233459 


Bearing  from  B  to  10-50  B          =  N.W.  37°  36'  27"      tan.  =  9-8836647 

Station  10-50    Bx  co  dx    =    5873-969  co  6!21'610  =  255-641  log.  =  2-4076305 

BE  co  dE     =      264-881  co      53-933  —  210'916  log.  =  2-C241713 


Bearing  from  B  to  10-50  d          =  N.W.  39°  31'  41"      tnn.  =  9-9KJ5408 


Station  ll'OO    Bx  co  ax    =    5873-969  co  6117-100  =  273-131 
BE  co  aE    =      264-884  co      96-119  =  168'765 


log.  =  2-43G3710 
log.  =  2-2272824 


Bearing  from  B  to  11-00  a          =  N.W.  31°  42'  41"      tan.  =  9-7909114 

Station  11-00    Bx  co  fcx    =    5873-969  co  6146-970  =  273-001  log.  =  2-4361642 

BE  co  6E     =      264-884  co     104-165  =  160 '719          -      leg.  =  2-2060672 


Bearing  from  B  to  11-00  b          =  N.W.  30°  29'  09"      tan.  =  9'7699030 


132  USEFUL      FORMULAE. 

Station  11-00    Bx  co  ex     =    5873-969  co  6129'IOG  =  255-131  log.  =  2-4067632 

BE  co  CE    =      264-884  co      95 "838  =  169-046  Jog.  =  2-2280049 


Bearing  from  B  to  11 -00  b          =  N.W.  33=  31'  40"      tan.  =  9-8212417 

Station  11-00    Bx  co  ds    —    5873-969  co  6128-970  =  255-001  log.  =  2-4065419 

BE  co  dE     =      264-881  co     103-860  =  161-024  log.  =  2-2C68906 


Bearing  from  B  to  11-CO  d          =  N.W.  32°  16'  15"      tan.  =  9-8003487 

Station  11-50    Bx  co  «x    =    5873-969  co  6146-110  =  272-141  log.  =  2-4347940 

BE  co  OE     =      264-884  co    146-183  =  118-701  log.  =  2-0744544 


Bearing  from  B  to  11-50  a          —  N.W.  23O  £3'  56"      tan.  =  9-C3S6604 

Station  11-50    Bx  ^  ^     =    5873-969  <*=  6145-920  =  271-951  log.  =  2-4344907 

BE  co  ^E    =      264-884  co    154-223  =  110-656  log.  =  2-0439751 


Bearing  from  B  to  11-50  b          =  N.W.  22°  08'  29"      tan.  —  9-6194844 

Station  11-50    Bx  ^  ex     =    5373-969  co  6123-120  —  254-151  log.  =r  2-4050918 

BE  co  CE     =      264-884  co    145-755  =  119-129  log.  =  2-0760175 


Bearing  from  B  to  11-50  e          =  N.W.  23°  06'  51"      tan.  =  9'6709257 

Station  11-50    Bx  co  rfx    =    5873-969  co  6127-920  =  253-951  log.  =  2*4047499 

BE  co  dE     =      264-884  co    153-176  =  111-708  log.  =  2-0480883 


Bearing  from  B  to  11-50  d          =  N.W.  23^  44'  38"      tan.  =  9-6433384 

Station  12 '00    Bx  co  ax    =    5783-969  co  6144-720  =  270-751  log.  =  2-4325701 

BE  co  OE     =      264-884  co     196-237  =    68-647  log.  =  1-8366216 


Bearing  from  B  to  12-00  a          =  N.W.  14°  13'  38''      tan.  =  9-4040515 

Station  12-00    Bx  ^  b*     =    5783-969  co  6144-450  =  270*481  log.  =  2-4321368 

BE  co  br.     =      264-884  co    204-280  =    60-604  log.  =  1-7825013 


Bearing  from  B  to  12 -00  b          =  N.W.  12°  37'  45"      tan.  =  9-35C3645 

Station  12-00     Bx  ««  ex     =    5873-969  «~  6126-730  =  252-761  log.  =  2'4C27100 

BE  co  CE     —      264-884  co     195-662  =    69'222  log.  =  1-8402441 


Bearing  from  B  to  12-00  c          =  N.W.  15°  18'  5C"       tan.  =  9 -43753-11 


BEARINGS      OF      STATIONS. 

Station  12 '00    Bx  <*>  dx    =    5873-969  <*>  6126-460  =  252 '491  log.  =  2-4022459 

BE  co  dE    =      264-884  <*>    203-682  =    61*202  log.  =  1-7867656 

Bearing  from  B  to  12-00  d          =  N.W.  13°  37'  31"  tan.  =  9-3845197 

Station  12-50    Bx  v=  a*     =    5873-969  <*>  6142-920  =s  268-951  log.  =  2'4296731 

BE  co  aE     =      264-884  co    246-177  =    18-707  log.  =  1-2720041 

Bearing  from  B  to  12-50  a         =  N.W.  3°  58'  44"  tan.  =  8-8423310 

Station  12'50    BN  <~  by     =    5873-969  co  6142'590  =  268-621  log.  =  2-4291399 

BE  co  &E     =      264-884  ^    254-318  =    10-566  log.  =  1-0239106 

Bearing  from  B  to  12-50  b          =  N.W.  2°  15'  09"  tan.  =  8'5947707 


133 


Station  12-50    Bx  co  ex     =    5873-969  co  6124-930  =  250-961 
BE  co  CE     =      264-884  co    245-557  =    19-327 


log.  r=  2-3996062 
log.  =  1-2861644 


Bearing  from  B  to  12-50  c          =  N.W.  4°  24'  13"         tan.  =  8-8865582 


Station  12-50  Bx  <*>  dx  =  5873-969  co  6124'600  =  250*631 
BE  co  dE  =  264-884  co  253-574  =  11-310 
Bearing  from  B  to  12-50  d  =  N.W.  2°  35'  02"  tan.  =  8-6544278 


log.  =  2-3990348 
log.  =  1-0534626 


Station  13-00    BN  co  ax    =    5873-969  co  6140*790  =  266'821 
BE  co  aE    =      264-884  co    296-302  =    31-418 


log.  =  2-4262200 
log.  =  1-4971785 


Bearing  from  B  to  13 -00  a          =  N.E.  6°  42'  56"          tan.  =  9-0709585 


Station  13'00    BN  co  CN    =    5873*969  co  6122-810  =  248-841 
BE  co  CE    =      264-884  co    295-435  =    30-551 


log.  =  2-3959220 
log.  =  1-4844564 


Bearing  from  B  to  13*00  c          =  N.E.  6°  59'  25"          tan.  =  9-0885344 

Station  13-50    Bx  co  13'50x  =  5873-969  co  6128'870  =  254*901  Jog.  =  2-4063715 

BE  co  13-50E  =:    264  "884  co    349*811  =    84*927  log.  =  1*  9290458 

Bearing  from  B  to  station  13*50  =  N.E.  18°  25'  37"         tan.  =  9*5226743 

We  have  thus  completed  our  computations  of  bearings,  from  the 
points  A  and  B  to  the  corners  of  the  piers,  etc.  We  now  arrange 
them  in  the  following  table,  and  from  the  bearings  between  A  and 


134 


USEFUL      FOKMUL.E. 


B,  and  between  A  and  the  corner  of  the  piers,  etc.,  we  readily  deduce 
the  angles  required  to  be  measured  at  A,  and  in  like  manner  those 
required  to  be  measured  at  B. 

A      TABLE       OF      BEARINGS, 

FROM    STATIONS    A    AND     B    TO    THE     CORNERS    OF    THE     SEVERAL    PIERS    AND 
ABUTMENTS,  AND   TO   STATIONS   10  AND   13-50  ; 

WITH   THE    ANGLES    TO   MEASURE   FROM    A    AND    B  TO    EACH    CORNER   AND    STATION    INDICATED. 


PIERS 

AND 

CORNERS. 

BEARINGS  FROM 
•STATION  A  TO  THE 

POINTS    INDICATED 
IN   COLUMN    1. 

BEARINGS  FROM 
STATION  B  TO  THE 

POINTS   INDICATED 

IN  COLUMN  1. 

ANGLES 
AT  STATION  A 

WITH    B.    AND 
THE   POINTS 
INDICATED   IN 

COLUMN  1. 

ANGLES 
AT  STATION  B 

WITH   A,   AND 
THE   POINTS 

INDICATED  IN 
COLUMN  1. 

10-00  station 

0       /        /' 

N.E.     20   00   00 

0       /        // 

N.W.     45  00  00 

0        /       '/ 

83   00   00 

0        /        " 

32   00    00 

10-50          b 

"       31   54  51 

"         37  36  27 

71   05    09 

39   23   23 

d 
11-00          a 

"        34  23  49 
"       39   59   00 

'•         39   31  41 

"         31  42  41 

68   36    11 
63   01   00 

37  23    19 
45   18   19 

"             b 

"       41   21   15 

"         30   29   09 

61  38  45 

46   30  51 

"             c 

"        42  40   29 

"         33  31  40 

60   19  31 

43  28  20 

"             d 

"        44   03    10 

"         32   16   15 

58  56  50 

44  43  45 

11-50           a 

"       47  52  48 

"         23  33  56 

55   07   12 

53  26   04 

"              6 

"       48  47  04 

"          22  08   29 

54   12  56 

54  51   31 

"              c 

"       50  23  38 

"          25   06   51 

52   £6   22 

51   53   09 

"               c/ 

"       51   22  56 

"         23  44  38 

51   37   04 

53   15   22 

32-00          a 

"        53  47  00 

"          14   13  38 

49    13   00 

62  46   22 

"              6 

"        54  38  06 

"          12  37  45 

48  21   54 

C4  22    15 

"              c 

"       56  20  39 

"         15   18  56 

46   39   21 

Cl   41    04 

"              d 

"        57  09  04 

"          13  37  31 

45  50  56 

63   22  29 

12-50          a 

"       58   35  37 

"           3  58  44 

44  24  23 

73   01    16 

// 

"       59   17  33 

"            2    15   09 

43  42   27 

74  44  51 

"              c 

"        61   00   08 

"            4   24   13 

41   59   52 

72  35  47 

d 

"        61   39   22 

"            2   35   02 

41   20   38 

74  24   58 

13-00           a 

"        62  29   50 

N.E.         6  42  56 

40   30    10 

83  42   56 

c 

"       64  42  41 

"           6   59   25 

38    17    19 

83  59    25 

13-50  station 

"       66   57  01 

"          18  25  37 

36   02   59 

95  25   37 

A 

N.W.     77  00  00 

B 

S.E.      77  00  00 

DIFFERING     GRADES     AND     VERTICAL     CURVES.      135 

We  have  thus  completed  our  table  of  angles  which  are  to  be  used 
in  the  location  of  the  points  indicated  as  follows.  Having  two 
observers  with  instruments,  one  at  station  A  and  the  other  at  B? 
each  having  a  suitable  instrument,  they  proceed  to  lay  off  upon 
their  respective  instruments  the  angles  indicated  in  the  table  to 
any  one  of  the  corners  or  stations  desired.  Having  done  this,  an 
assistant  repairs  with  a  boat  to  the  place  of  intersection  of  the  lines 
corresponding  with  their  instruments  ;  and,  if  the  water  be  not  too 
deep,  fixes  a  stake  by  driving  it  into  the  mud  or  sand  which  forms 
the  bottom  of  the  river.  Or,  if  piles  are  being  driven  from  a  scow, 
the  position  of  the  pile  may  be  brought  to  the  intersection  indicated 
by  the  instruments,  and  driven.  Or,  the  point  may  be  otherwise 
marked  by  mooring  a  buoy  or  float  by  the  aid  of  two  or  three  lines ; 
and  doubtless,  by  many  other  devices,  marks  may  be  fixed  which 
will  be  found  equally  simple  and  exact,  the  whole  of  the  mechanical 
operations  being  so  simple  in  their  character  as  not  to  need  further 
description.  I  will  only  add  that  we  have  made  use  of  the  method 
which  we  have  here  endeavored  to  develope  in  several  instances,  and 
have  found  it  very  convenient  and  accurate. 

(51)  We  have  thus,  in  the  foregoing  pages,  completed  our 
contemplated  essays  upon  railroad  curves  connected  with  the  aline- 
ments  of  the  main  tracks,  side  tracks,  and  turnouts;  we  now 
propose  to  add  a  formula  for  uniting  the  different  gradients  of 
railroad  tracks  with  vertical  curves. 

Before  we  proceed  to  the  investigation  of  formula,  we  would 
remark,  it  is  not  our  purpose  to  give  anything  like  a  full  description 
of  the  operations  for  laying  down  the  gradients  of  a  railroad 
track,  (the  operations  being  so  simple  in  character  as  to  be  readily 


[FiG.     21.] 


VERTICAL      CURVES.  137 

comprehended  by  every  one,)  but  merely  to  develope  a  formula 
which  has  been  found  convenient  and  useful  in  our  practice  of 
rounding  off  the  salient  angles  and  hollowing  the  re-entering  angles 
(if  they  may  be  so  termed)  formed  by  the  intersections  of  the 
gradients  of  a  railroad  track. 

(52)  In  our  practice  we  have  never  laid  down  a  vertical  curve 
of  a  less  radius  than  forty  thousand  feet,  but  in  general  our  curves 
have  embraced  two  hundred  feet  upon  either  side  of  the  intersecting 
point  of  the  gradients ;  that  is  to  say,  the  vertical  arc  has  usually 
been  about  400  feet  in  length  ;  but,  when  the  inclinations  of  the 
gradients  have  been  such  as  to  make  the  angles  to  be  rounded 
or  hollowed,  comparatively  acute,  we  have  sometimes  used  a 
shorter  arc. 

Presuming  that  the  inclinations  of  the  gradients  and  the  relative 
positions  of  the  angles  have  been  determined,  we  commence  with 
the  investigation  of  formula  for  determining  the  value  of  these 
angles  in  degrees. 

The  problem  presents  four  different  cases,  viz.,  the  salient  angle 
formed  by  an  ascending  and  descending  grade.  The  re-entering 
angles  formed  by  two  descending  or  ascending  grades,  one  of  which 
being  much  more  inclined  than  the  other.  The  re-entering  angle 
formed  by  a  level  line,  and  one  ascending  or  descending  grade.  The 
re-entering  angle  formed  by  a  descending  and  an  ascending  grade. 

To  explain  the  foregoing  angles,  and  the  method  of  ascertaining 
their  value  in  degrees  and  minutes,  (See  Fig.  21,)  let  a  a  represent 

a  level  line,  A  S  an  ascending  grade,  and  S  D  a  descending  grade, 
T 


138  USEFUL      FORMULA. 

then  will  the  angle  at  S  be  a  salient  angle.  To  determine  the 
measure  of  this  angle,  viz.,  A  S  D,  we  will  suppose  A  S  to  ascend 
at  the  rate  of  forty  feet  to  the  mile,  and  S  D  to  descend  at  the  rate 
of  twenty-five  feet  to  the  mile.  We  will  now  suppose  S  a  to  equal 
one  mile,  or  5280  feet;  S  D  and  S  A'  are  also  taken  as  a  mile  each; 
inasmuch  as  there  will  be  no  practical  difference  between  the 
length  of  a  line  inclining  40  or  even  80  feet  to  the  mile,  and  the 
same  line  reduced  to  a  level,  (and  this  remark  will  apply  almost 
universally,  or  to  the  gradients  of  railroads  in  general :)  therefore 
the  lines  S  A,  S  a,  S  D',  are  taken  each  as  one  mile. 

From  an  inspection  of  the  figure,  it  will  be  obvious  that  the  angle 
A  S  D  will  equal  180  —  (A  S  a  +  D  S  a')  and  the  angle  A  S  a  = 
A'  S  a ' ;  wherefore,  A  S  a  -}~  ^  S  a'  =  D  S  A' ;  an(j  ^e  f0nowing 
method  of  determining  which,  though  not  strictly  accurate,  will  be 
found  sufficiently  exact  for  every  practical  purpose.  Taking  S  D 
and  S  A'  =  one  mile  each,  then  will  a  A'  =  40  feet,  and  D  a  = 
25  feet,  and  the  angles  S  D  A'  and  S  A'  D  being  each  so  near  a 
right  angle  that  we  may  take  either  of  them  as  such.  Taking  the 
angle  D  as  a  right  angle,  S  D  will  be  a  cosine,  and  D  A'  a  sine, 
and  S  A'  will  be  a  radius. 

Then,  taking  radius  =  unity,  and  representing  it  by  E,  we  have 
this  analogy,  cos.  :  sin.  ::  R  :  tan.  =  ~-  ;  which,  in  practice,  will 
stand  thus, 

S  D  :  (I)  a  +  a'  A')  : :  R  :  tan.  A'  S  D  =  (D  a/sV  A°  5 
and  180°  — A'  SD  =  ASD; 

and  consequently,  the  angle  A'  S  D  =  C  in  the  quadrilateral 
C  T  S  T.  (63) 


VERTICAL      CURVES.  139 

Having  obtained  the  angles  S  and  C  in  the  quadrilateral,  we  will 
now  proceed  to  give  an  example  of  computation.  As  before  stated, 
(D  a  +  a'  A)  =  D  A'  =  25  +  40  =  65  feet,  and  S  D  =  5280 
feet  =  1  mile;  then,  we  have 

D  A'  =    65      feet        log.  =   1-8129134 
S  D    =    5280     "         log.    =  3-7226339 


C         =    0°  42'  19"     tan.  =  8-0902795 

We  take  the  angle  C  to  the  nearest  second  ;  it  is  not  necessary 
that  we  should  be  more  exact. 

Then,  taking  the  distances  S  T  and  S  T'  =  260  feet  each,  we 
ascertain  the  radius,  (in  our  computations  it  is  not  necessary  to 
know  the  radius,  and  we  merely  ascertain  it  this  time  as  a  matter 
of  curiosity  rather  than  use ;)  to  find  which,  we  have 

Sin.  i  C  :  260  feet  : :  cos.  |-  C  :  radius  =  cot,  |  C  260      (64) 

Thus,  260  log.  =  2-4149733 

£  C  =  00  21'  10"  cot.  =  2-2106159 


Radius  =  42226  feet       log.  =  4-6255892 

We  have  thus  found  our  radius  =  42226  feet,  the  angle  S  being 
very  acute,  (speaking  comparatively,)  it  becomes  necessary  to  take 
the  distances  S  T  and  S  T'  somewhat  greater  than  it  is  our  habit. 
We  would  remark  here,  that  whatever  distance  we  assume  for  S  T, 
it  will  be  found  convenient  that  it  should  divide  even  by  the 
number  20,  because  in  setting  the  grade-pins  for  laying  down  the 
rails,  it  is  usual  to  place  them  20  feet  apart,  which  is  as  long  as 
we  can  conveniently  have  the  straight-edged  board  used  as  a  guide 
in  placing  the  sleepers  or  ties  to  the  proper  height  or  grade. 

We  have  taken  S  T  =  260  feet,  which,  divided  by  20  feet,  will 


[Fia.     22.] 


VERTICAL      CURVES.  141 

give  thirteen  divisions  upon  each  side  of  S;  consequently,  the  portion 
of  the  arc  spanned  by  one  of  these  divisions  will  be  equal  to-—-. 

We  have  demonstrated  in  section  (2)  that  the  angle  of  deflection 
is  equal  to  half  the  centre  angle,  spanned  by  the  chord  governing 
said  deflection.  We  therefore  have  in  the  large  triangle  TCI,  the 
angle  at  T  =  90,  consequently  the  angle  at  1  is  equal  to  the 
complement  of  the  angle  C  ;  and  the  angle  C  =  ^-~p-  =  -*  ^  1C 
=  0°  Or  37" -7;  hence  the  angle  at  1  =  89°  58'  22" -30,  which 
is  common  to  the  triangles  TIC  and  T  I  a;  then,  in  the  triangle 
T  1  a,  we  have  T=  °°  0!/237"-70  -=-  Q°  00'  48"  '85;  hence  the  angle 
at  a  will  be  equal  to  180°  —  (89°  58'  22"  -30  +  00'  48"  -85)  = 
90°  00'  48" -85.  That  is,  the  angle  at  a  is  equal  to  90°  +  the 
angle  at  T ;  and  the  angle  b,  in  the  triangle  T  b  2,  will  also  be 
equal  to  90°  +  the  angle  at  T;  and  the  same  remark  will  apply  to 
the  triangles  T  c  3,  T  d  4,  etc.,  to  T  m  S. 

Having  thus  explained  the  method  of  deducing  the  angles,  we 
find  the  ordinates  a  1,  62,  c  3,  etc.,  to  m  S,  by  the  following  formula. 
Taking  the  triangle  T  a  1  for  an  example,  and  making  use  of  the 
symbols  belonging  to  the  same,  we  have  sin.  90°  -f-  T  I  T  1  : :  sin. 
T  :  a  I ;  but,  since  the  sin.  of  90°  +  T  is  equal  to  the  cos.  T, 
the  formula  may  be  rendered  thus,  cos.  T  :  T  1  : :  sin.  T  :  a  1  = 
tan.  T  X  T  1  (65) 

We  wish  to  remark,  that  in  the  investigation  above,  we  have 
considered  the  arc  T  m  T  and  the  tangent  lines  T  S  and  S  T 
of  equal  length ;  consequently,  the  computed  ordinates  will  be 
practically  the  same  as  if  they  were  perpendiculars  to  the  tangent 
line  T  S. 


1-t-  USEFUL      FORMUUB. 

We  will  now  show,  by  actual  computations,  that  the  formula, 
though  not  in  the  strictest  sense  exact,  still  it  presents  by  far  a 
greater  degree  of  accuracy  than  would  be  possible  to  practise  in 
laying  down  railroad  tracks. 

As  a  test  to  our  formula,  we  will  now  determine  the  length  of 
the  curve  T'  m  T,  and  compare  the  same  with  the  tangent  lines 
T  S,  S  T'.  In  the  triangle  S  T  C,  we  have 

Sin.  C  :  S  T  : :  cos.  C  :   rad.  =  cot.  C  X  S  T. 

Then,  representing  S  T  by  £,  the  radius  in  seconds  by  r" ,  and  the 
angle  C  in  seconds  by  C",  and  the  radius  in  the  unit  of  measure 
(found  above)  by  7%  and  the  length  of  the  curve  by  c,  we  shall  have 

r  .  C"  cot.  C  .  t  .  C" 


Thus,  C     =  0°  21'  00"  cot.  =  2-2106159240 

t      =    260  feet  log.  =  2 '4149733480 

C"  =  2540    "  log.  =  3-4048337166 

r"    =  co.  ar.  log.  =  4-6855748668 


c      =    519-9934  feet         log.  =  2-7159978554 
2t   =    520 


Difference  =        0'0066 

We  find  thus,  that  the  arc  and  the  tangent  lines  agree  within 
10060"  °f  a  f°°t'  whi^  is  a  little  larger  than  ^  of  an  inch,  a 
quantity  quite  too  small  to  be  considered  an  error  in  laying  down 
a  railroaj  track,  especially  in  the  ordinates  where  the  error  in  the 
greatest  will  be  reduced  in  the  proportion  the  length  of  radius 
bears  to  the  ordinate,  which  can  never  amount  to  any  thing -worth 
noticing,  especially  when  we  consider  that  the  case  we  are  examining 
is  of  that  class  which  produces  errors  greater  in  amount  than  the 
most  of  cases  which  come  under  consideration ;  so  that  I  think  we 
may  be  warranted  in  pronouncing  our  formula  practically  exact. 


VERTICAL      CURVES. 


143 


Having  investigated  the  necessary  formula,  we  will  now  proceed 
to  give  a  specimen  of  calculation. 

We  have  found  in  the  foregoing,  the  angle  T  in  the  triangle  T  a 1 
=  00'  48"  -85,  and  in  the  triangle  T  b  2  it  will  be  twice  that  amount, 
and  in  the  triangle  T  c  3  it  will  be  three  times  that  amount;  and  so 
on  to  the  centre  ordinate  m  S,  which  will  be  thirteen  times  the 
amount. 


Commencing  with  T  a  1      we  have    T 

T  1 
a  I 


0°  00'  48" -85  tan.  =  6  "3744395 

20  feet  log.  =  1-3010300 

0-00473663  feet        log.  =  7-6754695 


In  the  triangle    T  ft  2      we  have    T 

T  2 
ft  2 


QO  oi'  37" -07  tan.  =  6-6754695 

40  feet  log.  =  1-6020600 

0-0189465  feet         log.  =  8-2775295 


In  the  triangle    T  c  3      we  have    T 

T  3 
c  3 


0°  02'  26" -55  tan.  =  6-8515608 

60  feet  log.  =  1-7781513 

0-0426297  feet         log.  =  8-6297121 


In  the  triangle     T  d  4      we  have    T 
T  4 
d  4 


0°  03'  15"-40  tan.  =  6-9764996 

80  feet  log.  =  1-9030900 

0-0757861  feet          log.  =  8-8795896 


In  the  triangle     T  e 


we  have    T 
T  5 
e  5 


00  04'  04" -25 
100  feet 
0-118415  feet 


tan.  =  7-0734097 
log.  =  2-0000000 


log.  =  9-0734097 


In  the  triangle     T/6      we  have    T  —  0°  04'  53"-01  tan.  =  7-1525910 

T  6        =120  feet  log.  =  2-0791812 

/  6        =      0-170518  feet  log.  =  9-2317722 


In  the  triangle     T  g  7      we  have    T  —  0°  05'  41"- 95  tan.  =  7-2195379 

T  7        =140  feet  log.  =  2-1461280 

gl         =      0-232095  feet          log.  =  9-3656659 


144 


USEFUL      FORMULAE. 


In  the  triangle     T  h  8      we  have    T 

T  8 
h  8 

In  the  triangle     T  i   9      we  have    T 
T  9 
i  9 


:  0°  06'  30" -08  tan.  =  7-2775300 

:  160  feet  log.  =  2-2041200 

0-303144  feet  log.  =  9-4816500 

0°  07'  19"- 65  tan.  =  7-3286827 

180  feet  log.  =  2-2552725 

0-383668  feet  log.  =  9-5839552 


In  the  triangle     T  j  10    we  have    T  =0°  08'  08"'50 

T  10      =  200  feet 


tan.  =  7-3744403 
log.  =  2-3010300 


j    10      =      0-473664  feet          log.  =  9-0754703 


In  the  triangle     T  k  11    we  have    T  =0°  08'  57"-35 

T  11      =  220  feet 


tan.  =  7-41583S2 
log.  —  2'3!2'227 


k    11      =      0-573134  feet  log.  =  9-7562559 

In  the  triangle     T  /  12    we  have    T  =  0°  09'  46" -20  tan.  =  7 '4536218 

T  12      =  240  feet  log.  =  2-38(2112 


12      =      0-682076  feet  log.  =  9-8338330 


In  the  triangle     T  m  S    we  have    T  =  0°  10'  35" -05  tan.  =  7-4883842 

T  S       =260  feet  log.  =  2-4149733 

m  S       =      0-800493  feet          log.  =  9- 


We  have  thus  computed  the  thirteen  ordinates  according  to  the 
formula.  It  will  be  seen  that  the  first  angle  in  the  triangle  T  a  1 
is  taken  a  very  small  amount  too  large,  which  will  make  all  the 
angles  used  something  large,  hut  not  sufficiently  so  as  to  practically 
affect  our  results. 


The  reason  why  we  did  not  correct  the  angles  in  the  course  of 
our  operations  was,  that  we  may  he  enabled  to  compare  the  results 
obtained  by  another  method,  which  will  much  abridge  the  work; 
and,  although  not  strictly  accurate,  still  we  may  state,  as  we  have 
before,  respecting  the  preceding  formula,  that  it  is  practically 
exact. 


VERTICAL      CURVES. 


145 


The  other  method  may  be  explained  thus.  Having  divided  the 
line  T  S,  in  a  suitable  number  of  parts,  which,  for  the  purpose  of 
comparing  with  our  previous  computations,  we  will  suppose  to  be 
thirteen,  of  20  feet  each ;  we  must  then  compute  the  first  ordinate, 
viz.,  a  1,  in  the  triangle  T  a  1,  which  we  call?/;  which,  however, 
in  the  present  case  will  be  needless,  as  we  have  it  already  computed. 
We  therefore  take  the  value  of  y  from  our  previous  computations, 
which  of  course  will  need  no  comparison ;  we  then  find  the  re 
mainder  of  the  ordinates,  b  2,  c  3,  d  4,  etc.,  to  m  S,  according  to 
the  expressions  given  in  the  following  table. 


COLUMN 

OF 

ORDINATES. 

EXPRESSION 
OP                             COMPUTED 
FORMULA.                         RESULTS. 

COMPUTED  RESULTS 
BY  PREVIOUS  FORMULA, 
FOR  COMPARISON. 

No.  1. 

No.  2.                         No.  3. 

No.  4. 

No. 

1   or  a    1 

=         y           =    0-00473663  feet 

0-00473663 

feet 

2 

"62 

=     22  y             —     0-01894652     " 

0-01894650 

u 

3 

"     c     3 

=     32  y             =    0-04262967     « 

0-0426297 

(( 

4 

"     d    4 

=    42  y             =     0-07578608     « 

0-0757861 

({ 

5 

"     e     5 

=     52  y            =     0-11841575     " 

0-1184150 

« 

6 

"    /    6 

=     62  y             —     0-17051868     " 

0-1705180 

u 

7 

"     <7    7 

=     72  y             —     0-23209487     " 

0-2320950 

« 

8 

"     h     8 

=     82  y             =     0-30314432     " 

0-303145 

u 

9 

"     i     9 

=     92  y             =     0-38366703     " 

0-383668 

u 

10 

"   j  10 

=  102  y             =     0-47366300     " 

0-473664 

u 

11 

"     k  11 

=  U2  y             =     0-57313223     " 

0-573134 

« 

12 

"     Z  12 

=  122  y             =     0-68207472     " 

0-682076 

1C 

13 

"     m  S 

=  132  y             =     0-80049047     " 

0-800493 

K«7 

EXPLANATION  OF  THE  TABLES.  The  first  column  contains  the 
number  of  the  ordinates,  arranged  in  numerical  order,  from  one  to 
thirteen.  The  second  column  contains  tjbe  notation  expressing  the 


146  USEFUL      FORMULAE. 

method  of  computation.  Third  column  contains  the  resulting  com 
putations.  Fourth  column  contains  the  same  elements,  computed 
"by  the  preceding  formula,  which  is  placed  here  for  the  purpose  of 
conveniently  comparing  the  results  of  the  two  methods. 

The  comparison  shows,  that  the  method  by  squares,  is  sufficiently 
accurate  for  the  most  exact  work,  when  we  consider  we  can  only 
make  use  of  the  three  first  decimals,  in  practice,  while  the  two 
methods  do  not  differ  at  all  in  the  fifth  decimal.  I  repeat,  we  may 
without  hesitation,  pronounce  the  rule  practically  exact. 

(53)  In  applying  the  foregoing  results   to  practice,  doubtless 
different  engineers  will  pursue  different  methods;  but,  a  convenient 
method  is  to  ascertain   the  total  heights   (as   they   are  generally 
called)  of  stations  in  the  inclined  lines,  which  shall  correspond  to 
stations   of  the  same  number,   belonging  to  the   vertical  curves ; 
then  adding  or  subtracting  the  computed  elements  of  the  curve,  or 
length  of  ordinates,  corresponding  to  the  station  set  out  in   the 
inclined  lines,  accordingly  as  the  nature  of  the  case  requires. 

(54)  It  sometimes  happens  that  the  locating  stations  are  ob 
literated,  and  in  that  case,  the  position  of  the  angle  at  S,  (which  is 
the  main  starting  point,)  cannot  be  readily  found.     In  such  cases  the 
engineer  must  measure  a  portion  of  the  road  anew,  extending  the 
measurement  sufficiently  far  from  the  apex  or  point  S,  upon  both 
sides,  so  as  to  ascertain  the  grades  correctly.     Having  completed  the 
measurements,  and  marked  and  numbered  the  stations,  (which  are 
usually  fixed  one  hundred  feet  apart,)  and  determined  their  relative 
levels,  and  the  inclination  of  the  grades  which  govern  our  operations, 
we  proceed  to  ascertain  the  points  of  intersection. 


VERTICAL      CURVES.  147 

First,  draw  the  lino  k  k'  to  represent  a  level,  (See  Fig.  23  (a);) 
then,  draw  the  line  n  I',  at  an  angle  with  k  k'  ,  equal  to  the  inclination 
of  the  grade  S  n.  Again,  through  the  point  S,  draw  the  line  n  I  at 
an  angle  with  k  k'  ,  equal  to  the  inclination  of  the  grade  S  n  .  From 
the  point  n  let  fall  the  perpendicular  n  m;  and  also,  from  the  point 
n't  let  fall  the  perpendicular  n  I  ,  until  it  intersects  the  line  n  I  . 
Then,  from  the  intersection  of  n  I  with  n  I'  ,  draw  the  line  I  m 
parallel  with  k  k  ;  and,  from  the  intersection  of  n  I  with  n  I,  draw 
the  line  I  m  parallel  with  k  k'  . 

It  will  now  be  obvious,  from  an  inspection  of  the  diagram,  that 
the  angle  n  S  k  is  equal  to  the  grade  or  inclination  of  n  S  ;  and  the 
angle  k  S  I  is  equal  to  the  grade  or  inclination  of  S  n  .     And  it  will 
also  be  obvious  that  the  angle  n   S  k'  will  be  equal  to  the  grade  or 
inclination  S  n  ;  and  the  angle  k  S  I  will  be  equal  to  the  grade 
or  inclination  S  n.     Then,  representing  by 
d,  the  distance  between  tho  stations,  (usually  100  feet;) 
</,  the  difference  in  heights  between  the  stations  in  grade  n  S  ; 
g  ,  "  "  "  "  "       n   S; 

n  and  n  ,  the  numbers  of  the  stations  at  the  points  they  represent  ; 
h,  the  height  of  station  n; 
h'  ,  the     "  "        n  ;  we  have 

d  '  g   '.:   n  co  n  '.  n  m  ; 
and  d  '.  g    \\  n  <=o  n  :  n  m 


Then,  by  substituting  p  for  n  m,  and  p  for  n  m  ;  5  for  n  S, 
and  <$'  for  n  S  ;  and,  subtracting  p  from  h,  (which  gives  the  height 
of  the  line  m  I  above  the  datum  line,  and  which  we  will  represent 
by  0';)  then,  subtracting  p  from  Ji  ,  (which  gives  the  height  of  m  I 
above  the  datum  line,  and  which  wo  will  represent  by  0,)  we  have 


2.992  7J1 


71  12.452 


4.452m 


[Fia.     23(6).] 


1=44:52 


VEETICAL      CURVES. 


149 


:  a  ::  (K-o)  :  y  =<*=£, 


We  will  suppose,  for  an  example  of  calculation,  station  n  to  be 
numbered  10,  and  station  n  to  be  numbered  20;  h,  to  be  10'567 
feet  above  the  datum  line,  and  li  to  be  7  '385  feet  above  the  datum 
line  ;  g  to  be  0*7575  feet,  and  g  to  be  =  0*3787  ;  and  both  grades 
to  be  ascending  from  S ;  with  d  =  100  feet.  We  then  have 


Again, 


n  m  =  p 
h, 

p  —  h  =  o' 
d 

n  oo  n' 


=  100  feet  co. 

=  0-7575  feet 

=  10  stations,  or  1000  feet 

=      7-575  feet 
=    10-567 


log.  =  8-0000000 
log.  =  9-8793826 
log.  =  3-0000000 


2-992 

=  100  feet  co 

=      0-3787  feet 

=  10  stations,  or  1000  feet 


log.  =  0-8793826 


log.  =  8-0000000 
log.  =  9-5782953 
log.  =  3-0000000 


»' 

m'  =  p' 

=  3-787  feet 

V 

=  7-385 

It: 

—  p  =  o 

=  3-59H 

g* 

+  S 

=     1-1362  feet 

d 

=  100  feet 

ft 

—  0 

=      6-969 

log.  =  0-5782953 


log.  =  9-9445452 
log.  =  2-0000000 
log.  =  0-8431705 


<r  +  <r' 
a 

h  —  o' 


=      6-1336 

=  1-1362 
=  100  feet 
=  4-395 


log.  =  2-7877157 

log.  =  2-9445452 
log.  =  2-0000000 
log.  =  0-6427612 


log.  =  2-5873064 


Having  thus  ascertained  the  distances  n  S,  and  n  S  =  <5  and  <T  ; 
if  we  now  take  n  +  5  =  the  number  of  the  station,  represented 
by  S,  (the  point  of  the  intersection  of  the  grades  ;)  then  will  n  —  <5 
=  the  same  number  S,  if  the  computations  be  correctly  prepared. 


150  USEFUL  FORMULAE. 

STATIONS.  STATIONS. 

Thus,  n  —  10  n'  =  20 

f5  =    6-1336  f5'  =    3-8664 


—  16-1336  n'  —  6'   =  16-1336 


We  may  now  further  prove  our  work  by  ascertaining  the  height 
of  the  point  S,  from  the  datum  line,  by  computing  the  descent 
from  n  to  S,  and  from  ri  to  S.  If  our  computations  are  correct,  the 
results  should  be  alike.  Thus,  we  have  h  —  8.g  =  the  height  of 
S  ;  and  h'  --  S'.g'  =  height  of  S. 


6 

=    6-1336 

log. 

=  0-7877155 

8 

=    0-7575 

log. 

=  9-8793826 

6>g 

=    4-6462 

0-6670981 

h 

=  10-567 

S 

=     5-9208    above  datum  line. 

<y 

=    3-8664 

log. 

=  0-5873068 

s1 

=    0-3787 

log. 

=  9-5782953 

Height  of 


fj'.ff7          =     1-4642  0-1656021 

h1  —     7-385 

Height  of  S  =    5'9208    above  datum  line. 

Having  thus  found  the  station  corresponding  to  the  intersecting 
point  of  the  grades,  and  its  relative  height,  the  necessary  stations 
for  laying  down  the  vertical  curves  can  be  readily  prepared,  and 
the  work  can  be  proceeded  with  in  the  manner  set  forth  in  the 
foregoing. 

To  make  the  formula  just  enunciated  applicable  to  every  case 
would  require  several  modifications ;  we  shall,  however,  only  give 
one,  believing  that  the  ingenuity  of  the  reader  will  readily  supply 
whatever  may  be  deficient. 

(55)  The  case  we  propose,  is,  when  we  have   one   grade  de- 


VERTICAL      CURVES.  151 

scending  say  O'SOO  feet  per  100  feet,  which  will  intersect  another 
descending  grade  of  0*400  feet  per  100  feet. 

To  describe  the  construction  of  a  figure  applicable  to  this  case  we 
have  only  to  cop}r  verbatim  the  description  of  Fig.  23  (a);  we  there 
fore  refer  to  that  description  as  a  substitute. 

After  having  constructed  the  figure,  it  will  be  obvious  that  we 
have  but  few  modifications  to  make  in  the  formula  already  given ; 
but,  lest  we  should  not  be  fully  understood,  we  repeat  our  former 
formula,  with  the  necessary  modifications.  Thus  we  have 

d  :  g  : :  n  ^  ri  :  n  m;  and  d  :  g    : :  n  ^  ri  :  ri  m       (70) 

Then,  as  before,  substituting  p  for  n  m,  and  p  for  ri  m  ;  d  for 
n  S,  and  <$'  for  n  S ;  and  then,  subtracting  p  from  h,  we  get  the 
height  of  m  I'  above  the  datum  line,  (which  height  we  represent  by 
o  ;)  and  then,  by  adding  p  to  li  we  obtain  the  height  of  m  I  above 
the  datum  line,  (which  height  we  represent  by  o.)  We  then  have 

g  <^>  g  :  d  : :  h  —  o  :  5 
and  g  ^  g   :  d  : :  //,'  —  o   :  X  (71) 

EXAMPLE      OF      COMPUTATION. 

Let  d  =  100  feet ;  g  =  0-800  feet ;  g  =  0-400  feet ;  and  n  = 
10;  ri  =  20;  h  =12-452;  li  =  8 -234.  We  then  have 

Firstly,  d:  g  ::  n  <**  ri  '-  n  m  ; 
Secondly,  d  '.  g  '. '.  n  <^>  n'  *.  ri  m  . 

Thus,  d  =  100  feet  co.  ar.  log.  =  8-0000000 

S  =0-800  feet  log.  =  9-9030900 

n  w  n1         =10  stations,  or  1000  feet      log.  =  3-0000000 


n'  m  —p    =      8-00  feet  log.  =  0-9030900 


[Fia.     24.] 


2 3      4      5      B      7      8      9     ID 


6789     /lO 


CURVING      BOARDS.  153 

Secondly,        d                   =  100  feet  co.  ar.           log.  =  8-0000000 

^                   —      0-400  feet  log.  =  9-6020600 

n  •*>  ri         —  10  stations,  or  1000  feet  log.  =  3-0000000 

n'  m  =  p'    =      4-00  feet  log.  =  0-6020600 

h                          =  12-452  h'                         =    8-234 

p                          =    8-000  p'                         =    4-000 

(^  _  k)  —  Oi  —     4-452  (p1  +  A')  =  o  =  12-234 

h,'                        =    8-234  A                               12-452 


(k'—o1)           =    3-782  (h  —  0}              =    0-218 

Again,  we  have 

g  csz g  '  d  I',  h  —  o  :  8 ; 

and  g  ts*g'  '.  d  II  ft  —  d  '.  S                             ( 7 1 ) 

«•  oo  g-'    —      0-400  co.  ar.        log.  =z  0-3979400 

d               =  100  feet  log.  =  2-0000000 

h  —  0      =      0-218  feet  log.  =:  9-3384565 

,5               =  0-545  stations,  or  54-5  feet  log,  =  1-7363965 

§•  cx3  5-'    =      0-400  co.  ar.        log.  =;  0-3979400 

d               —  100  feet  log.  =  2 '0000000 

h'  —  o'     —      3-782  feet  log.  =  0-5777215 


<Y  =      945-5  feet,  or  9-455  stations  log.  5=  2-9756615 

(56)  We  conclude  our  remarks  upon  tracklaying  with  a  for 
mula  for  the  computation  of  elements  convenient  for  setting  out 
curving  boards  or  patterns  for  bending  rails  to  suit  the  horizontal 
curves  of  short  radii. 

The  principles  of  our.  present  formula  are  based  upon  a  chord 

of  the  arc  equal  in  length  to  the  longest  rails,  being  divided  into 

equal   spaces   or   abscissa ;   and   then   ascertaining  the   length   of 

the  corresponding  ordinates  which  shall  extend  therefrom  to  the 

V 


154  USEFUL      FORMULAS. 

periphery  or  curve.  Calculations  based  upon  strict  formula,  being 
somewhat  lengthy,  will  require  considerable  time  and  labor  to 
perform  them.  The  engineer  being  frequently  called  upon  to  give, 
in  great  haste,  the  elements  for  making  a  pattern  to  guide  the 
tracklayer  in  curving  his  rails,  it  becomes  desirable  to  obtain  a 
formula  as  short  and  convenient  as  practicable. 

These  considerations  have  led  to  the  adoption  of  the  following 
formula,  which,  though  not  strictly  correct,  is  nevertheless  as 
accurate  as  mechanical  skill  requires. 

By  way  of  explanation,  suppose  it  desirable  to  form  a  pattern  for 
bending  rails  of  twenty  feet  in  length,  it  will  be  found  convenient 
to  divide  the  chord  into  equal  parts  of  one  foot  each. 

From  an  examination  of  the  sketch,  it  will  be  obvious  that  one 
of  these  divisions  will  bisect  both  the  chord  and  the  arc,  and  that 
the  parts  thus  bisected  will  be  similar  and  equal ;  therefore,  the 
computations  made  for  the  one  part  will  apply  to  the  other. 

To  proceed  to  the  investigation,  we  first  ascertain  the  angle  at 
the  centre  of  the  curve  spanned  by  an  absciss  at  the  periphery  of 
one  foot. 

Representing  this  angle  by  C' ;  the  absciss  by  a ;  the  ordinatc 
corresponding  to  No.  1,  by?/;  and  the  radius  of  the  curve  by  r; 
then,  by  considering  r  a  cosine ;  and  the  absciss  a,  which  spans  the 
arc,  a  sine ;  we  have  the  following  analogy, 

Cos.  :  sin.  : :  R  :  tan.  C' ; 

which,  by  substituting  for  the  cosine  its  value  =  r;  and,  for  the 
sine,  its  value  =  a  =  unity  ;  we  then  have 


CURVING      BOARDS.  155 


Tan.  C  =  y-  and  cos.  \  C  :  a  \\  sin.  \  C  :  y  =  tan.  J  C'  .  a  = 
tan.  \  C'  .  1  ;    it  is  now  obvious  that  ----  =  2JL=  —  -  ;  wherefore, 

2  r  ' 

tan.  i  C  .  1  =#=  7  !a~  (72) 


Performing  the  computations  indicated,  by  logarithms,  we  have 
log.  y  =  (ar.  co.  log.  r  +  ar.  co.  log.  2.) 

It  will  be  seen  by  the  above  expression  that  we  have  considered 
the  arc  and  the  tangent  of  the  same  length,  which  will  be  found 
sufficiently  exact  for  every  practical  purpose,  and  that  the  ordinate 
represented  by  y  =  1  a,  as  represented  in  the  figure.  To  find  the 
remainder  of  the  ordinates  we  have,  for  ordinate 

No.  2  =  b  2  =  22  y 

3  =  c  3  =  32  y 

4  =  d  4  =  42  y 

5  =  e  5  =  52  y 

etc.,  to  the  number  of  ordinates  contained  in  half  the  chord  or 
tangent  line. 

For  the  purpose  of  testing  the  degree  of  accuracy  of  the  formula 
enunciated  above,  it  may  be  necessary  to  obtain  from  strict  com 
putations  the  ordinate  D  d  =  (10)  (10.)  This  ordinate  will 
correspond  to  102  ?/,  or  the  greatest  ordinate  of  the  computation, 
and  will  contain  a  greater  error  than  any  one  of  the  others. 

INVESTIGATION  OF  EXACT  FORMULA.   Let  r  represent  the  radius  of 

the  curve  ;  II  the  radius  of  the  tables  ;  \  ch  the  half  of  the  chord  AB  ; 

C  the  angle  D  C  B  ;  then  will  r  :  K  :  :  \  ch   :    sin.  C  =  ^-  ;  and 

Cos.  \  C  :  J  ch  :  :  sin.  }  C  :  D  d  =  tan.  \  C  .  J  ch  (74) 


156 


USEFUL     F  0  R  M  U  L  JE . 


which  expression  corresponds  to  the  greatest  ordinate  of  the 
calculations ;  and  which,  in  the  practical  examples  we  shall  give, 
will  be  represented  by  102  y.  So  that  the  difference  between  the 
ordinate  found  by  the  exact  formula  and  1C)2  y  will  constitute  the 
error. 

Example  of  computation,  according  to  the  approximate  formula. 
Assuming  r  =  oOO  feet ;  \  eh  =  10  feet ;  and  a  =  1  foot ;  we  shall 
then  have 


r      =  300  i'eet 
2       — 


0-0016666  feet 


co.  ar.        log.  1=  7-5228787 

co.  ar.        log.  =  9-6989700 

log.  =  7-2218487 


Having  found   !No.  1 

=  a 

1 

= 

y 

=  0- 

0016666 

we  have        2 

=  b 

2 

__  22 

y 

=  0- 

0066666 

3 

==  c 

3 

=  3- 

y 

=  0- 

0149999 

4 

=  d 

4 

=«i  & 

y 

=  0- 

0266666 

5 

=  e 

5 

=  5- 

y 

=  0- 

0416666 

6 

=  f 

6 

=  62 

y 

=  0- 

0599999 

7 

=  9 

7 

=  7- 

y 

=  0- 

0816666 

8 

=  h 

8 

=  8- 

y 

=  0- 

1066666 

9 

=  i 

9 

=  9- 

y 

=  0- 

1349999 

10 

J 

10 

=  10- 

y 

=  0- 

1666666 

Example  of  computation  of  the  greatest  ordinate,  according  to 
the  exact  formula,  as  a  test  to  the  above,  viz.,  by  formulae  (73)  and 
(74.)  We  have 


Sin.  C  = 


',  ch 


and  tan.  I  C  .  i  ch  =  1)  d  =  10  .  10 


r  =  300  feet 

=V  ch       =    10    " 

C  =  10  5V  36'' -767 


co.  ar.        log.  =  7-5288787 
log.  =  1-0000000 


sin.  =  8-5228787 


CURVING      BOARDS.  157 

C        =  00  57'  18" -383  tan.  =  8-2219693 

ch      =  10  feet  1-  0000000 


D  d       =  (10)  .  (10)  =  0-16671  feet  9-2219693 

10a  y     =  0-16666 


Difference  =  0-00005 

It  will  be  seen,  by  comparing  D  d,  with  102  y,  that  the  errors  of 
the  approximate  formula  are  too  small  to  be  noted  in  the  practical 
operations  of  curving  rails. 

It  will  be  noticed  that  the  above  ordinates  extend  from  the 
tangent  line  D  10,  D  10",  to  the  curve  A  B  D.  It  frequently 
happens  that  it  will  be  more  convenient  to  set  out  the  curve 
from  the  chord  line  A  d  B,  which  'line  may  be  readily  represented 
upon  the  pattern  board  by  straining  a  small  string  or  wire  from 
A  to  B.  To  prepare  ordinates  to  be  thus  used,  we  subtract  the 
ordinates  found  successively  from  the  greatest  ordinate,  (which,  in 
our  example,  is  ==  102  y.)  Thus, 


102  y 

— 

0 

=    D  d 

=     0-1666666 

102  y 

— 

y 

=     a 

V 

=     0- 

1650000 

I02  y 

—     22 

y 

=     b 

2' 

=     0- 

1500000 

102  y 

—     32 

y 

=    c 

3' 

—     0-1516666 

102  y 

—     42 

y 

=    d 

4' 

=     0- 

1400000 

102  y 

—     52 

y 

^^    e 

5' 

=     0- 

1250000 

102  y 

—     G2 

y 

=   f 

6' 

=     0- 

1066666 

102  # 

—     72 

y 

=    9 

7' 

=     0- 

0851111 

102y 

—     82 

y 

=    h 

8' 

=     0- 

0600000 

102  y 

—     92 

y 

=     i 

9' 

=     0- 

0316666 

102  ?/ 

—  102 

y 

=    J 

10' 

=     0- 

0000000 

Before  we  leave  this  subject,  we  would  remark,  that  all  curves 
of  a  less  radius  than  3000  feet,  provided  they  are  laid  with  rails 


158  USEFUL      FORMULAE. 

twenty-one  feet  in  length,  should  be  curved ;  and  a  board  prepared 
as  a  pattern  will  facilitate,  and  add  both  to  the  convenience  and 
accuracy  of  the  operations. 

(57)  It  frequently  becomes  necessary  for  the  engineer  to 
ascertain  the  radius  of  a  small  part  of  a  curve  in  a  railroad  track ; 
as,  for  example,  when  called  upon  to  lay  down  a  side  track,  a 
turnout,  or  to  connect  a  branch  road  with  a  curve,  the  radius  of 
which  is  unknown.  Many  formula  may  be  deduced  for  the  solution 
of  this  problem,  each  possessing  nearly  equal  convenience;  we  shall, 
however,  limit  our  investigations  to  some  two  or  three  of  those  in 
common  use. 

Let  T'  A  T  represent  a  portion  of  a  curve,  the  radius  of  which  it 
is  desirable  to  ascertain.  We  measure  the  chords  A  T  and  A  T', 
each  of  the  same  length,  which  we  represent  by  c;  then  ascertain 
the  point  B  in  the  middle  of  the  chord  T'  T;  and  measure  the 
ordinate  A  B,  which  we  represent  by  b.  Then,  representing  the 
radius  of  the  curve  by  r;  and  the  radius  of  the  tables  by  11 ;  we 
have,  in  the  triangle  A  B  T',  c  '.  R  : :  b  *.  cos.  A  =~(r  5  and 

For  an  example  of  computation,  we  will  suppose  c  =  100  feet; 
and  b  =  8  feet.  Then, 

c2  =  1002  log.  =  4-0000000 

b    =      8  co.  ar.  log.  =  9-0969100 

2  co.  ar.  log.  =  9-6989700 

r    =  625  feet  log.  =  2-7958800 

V 

Again,  let  T'  A  T  represent,  as  before,  a  portion  of  a  curve,  the 
radius  of  which  we  desire  to  ascertain  ;  let  the  chord  T'  T  be 


DETERMINATION      OF      RADIUS.  159 

represented  by  a;  and  let  a  =  199*359  feet;  and  6=8  feet; 
as  above.  We  then,  by  dividing  a  by  2,  have  in  the  triangle 
A  B  T,  (taking  b  as  a  cosine,  and  J  a  as  a  sine,)  the  following 
analogy, 

Cos.  .*  sin.  : :  R  :  tan.  A  =  -=p-  (7  6) 

Then,  Sin.  A  :  _ 

(77) 


XX 

A 

•!/»••   T?    •  <*•  —         ^  c 

-  =2 

*« 

.    2    "    •  •    -LI    •   /    . 

sin.  A  cos.  A 

EXAMPLE      OF     C  ALC 

U  L  ATIO  N. 

*0 

=  99-6795 

log. 

=  1-9986059 

b 

=    8                                co.  ar. 

log. 

=  9-0969100 

A 

=  850  24'  45"  -17 

tan. 

=  1-0955159 

A 

=  850  24'  45"  -17            co.  ar. 

sin. 

=  0'  00  13942 

A 

=                                    co.  ar. 

cos. 

=;  1'0969102 

2           co.  ar. 

log. 

==  9-6989700 

ka 

=    99-6795 

log. 

=r  1-9986059 

r 

=  625  feet                      radius 

log. 

=  2-7958803 

Again,  let  T'  A  T  represent  the  segment  of  the  curve  whose 
radius  is  desired,  and  the  half  chord  T'  B  ===  a;  the  ordinate  A  B 
=  b;  and  the  chord  T'  A  =  c;  and  the  diameter  A  D  =  d;  the 
radius  =  r.  We  here  remark,  that  the  angle  A  is  common  to 
the  triangle  T'  A  B,  and  the  triangle  T'  A  D ;  and  the  angle  B  in 
the  triangle  T'  B  A,  and  the  angle  T'  in  the  triangle  D  T'  A  are 
each  a  right  angle  ;  consequently,  the  two  triangles  are  similar. 
We  now  have 

a2  +  b2  =  cz 
and  b  :  c  : :  c  :  d  =  -£-  =    "'  j^  =  -£-  +  b  =  2  r         (78) 

EXAMPLE  OF  COMPUTATION.  Let  a  =  99-6795  feet;  b  =^  8  feet. 
Then, 


ELEVATING      OUTSIDE      RAIL      FOR      CURVE.         1G1 


a*          5=  99-G7952  feet 
b  =    8  feet 

1242 


log.  =  3-9972118 
co.  ar.   log.  —  9-0969100 

log.  =  3-0941218 


2)1250 


=   625 


We  have  here  thus  endeavored  to  apply  the  same  elements  in 
each  of  our  examples  of  computations,  by  way  of  testing  the 
different  methods,  and  we  find  each  of  them  to  give  the  same 
results. 

(58)  Before  we  take  our  final  leave  of  railroad  tracks,  we  will 
add  a  formula  for  elevating  the  outside  rail  of  curves.  We  extract 
what  we  shall  say  upon  the  subject  from  Article  XV.  of  the  22d 
vol.  of  the  American  Journal  of  Science,  1832. 

The  article  was  written  by  J.  Thompson,  (Engineer,  and  late  Pro 
fessor  of  Mathematics  in  the  University  of  Nashville,  Tenn.,)  and 
commences  with  a  discussion  of  a  formula  given  in  Colonel  Long's 
work  on  railroads,  which  he  shows  to  be  erroneous.  With  this 
criticism  it  is  not  our  intention  to  meddle ;  but,  as  Mr.  Thompson 
has,  in  the  course  of  his  remarks,  developed  a  convenient  and  an 
accurate  formula,  we  shall  endeavor  to  extract  only  so  much  as 
may  seem  to  be  connected  with  its  explanation. 


[w] 


162  USEFUL      FORMULA. 

"  Let  CAB  represent  a  horizontal  surface  on  which  a  railway  is 
situated ;  A  and  B  the  rails  placed  in  a  circular  curve  around  C  as 
a  centre.  A  car  moving  over  the  rails  A  and  B,  around  the  centre 
C,  will  be  acted  upon  by  two  forces,  one  horizontal  and  centrifugal, 
arising  from  the  motion  of  the  car  in  a  curved  line,  and  acting  in 
a  direction  from  the  centre  C ;  the  other,  the  force  of  gravity,  acting 
in  a  vertical  direction.  I  omit  here,  as  not  necessary  in  the  present 
investigation,  the  moving  force  derived  from  animal  or  other  power 
acting  in  a  direction  of  a  tangent  to  the  curve.  Let  the  horizontal 
line  A  K  represent  the  centrifugal  force  above  mentioned,  and  the 
line  E  A  the  force  of  gravity.  It  is  evident  that  the  resultant 
of  these  two  forces  will  be  E  K,  which  will  represent  both  the 
intensity  and  the  direction  of  the  pressure  of  the  loaded  car  upon 
the  rails.  The  line  E  K,  therefore,  representing  the  direction  of 
pressure,  the  rails  should  be  so  placed  that  this  line  may  be  perpen 
dicular  to  the  plane  passing  through  them.  Draw  the  vertical  line 
B  D,  and  through  A  draw  A  D,  perpendicular  to  E  K ;  B  D  will  be 
the  elevation  of  the  exterior  rail  above  the  interior,  and  the  angle 
DAB  will  be  the  inclination  of  the  plane  of  the  rails  to  the 
horizon.  The  centrifugal  force  A  K,  compared  with  the  force  of 
gravity  A  E,  is  easily  found,  when  the  radius  of  curvature  of  the 
track  and  the  velocity  of  the  car  are  given.  The  distance  between 
the  centre  C,  and  the  middle  of  the  track,  may  be  considered  as 
the  radius  of  curvature. 

"  We  may  obtain  a  very  simple  algebraic  expression  for  the 
elevation  of  the  exterior  rail.  Let  g  =  force  of  gravity;  c  = 
centrifugal  force ;  d  =  distance  between  the  rails ;  and  E  =  required 
elevation  ;  R,  and  V  representing  radius  and  velocity.  Then,  by  the 
similar  triangles  E  A  K  and  A  B  D,  we  have  E=  cd;  but,  by 


ELEVATING      OUTSIDE      KAIL      FOR      CURVE.          163 

central  forces,  c  =  -^-;  hence,  E  =  ~~-  in  this  expression;  g  is 
always  a  constant  quantity,  and  equal  to  32 -2  feet. 

"  If  the  velocity  of  a  car  on  a  railway  were  always  the  same,  we 
should  have  no  difficulty  in  assigning  the  proper  elevation  of  the 
exterior  rail.  But,  as  there  must  be  necessarily  a  great  variety  in 
rates  of  travelling,  an  elevation  for  a  rate  of  twenty  miles  per  hour 
would  he  much  too  great  for  a  rate  of  eight,  twelve,  or  fifteen  miles 
per  hour.  Perhaps  the  elevation  required  by  the  mean  velocity 
would  be  most  eligible.0  There  is  one  view  of  the  subject,  however, 
which  ought  to  be  taken  into  consideration  in  the  location  of  the 
exterior  rail.  When  a  car  moves  with  great  velocity  on  a  curved 
road,  and  the  planes  of  the  rails  are  horizontal,  the  flange  of  the 
fore  wheel  on  the  exterior  rail  is  exposed  to  very  great  friction, 
which  operates  as  a  retarding  force,  and  injures  both  the  car  and 
the  railway ;  this  friction  is  diminished,  though  not  altogether 
removed,  by  giving  the  exterior  rail  the  elevation  which  the  velocity 
and  radius  require.  In  order  to  reduce  the  friction  still  further,  or 
remove  it  altogether,  it  would  perhaps  be  advisable  to  increase  by 
a  small  quantity  the  elevation  obtained  as  above. f  It  is  evident 
that  a  car  moying  on  the  inclined  plane  A  D,  will  tend  by  its  own 
weight  to  approach  A,  and  recede  from  D  ;  this  will  oppose  the 
centrifugal  force  by  which  the  flange  is  pressed  against  the  rail  D, 
and  thus  the  friction  will  be  in  whole  or  in  part  removed.  I  know 
it  has  been  maintained  that  the  flange  of  the  hind  wheel  on  the 
interior  rail  produces  as  much  friction  as  the  flange  of  the  exterior 

*  It  has  been  the  practice  of  the  author  of  the  foregoing  papers  to  elevate  the  exterior  rail  to  suit 
the  highest  velocity  with  which  the  regular  trains  are  supposed  to  run  over  the  curve.  At  the  present 
time  we  should  not  think  thirty-five  miles  the  hour  too  great.  Quick  trains  produce  greater  friction 
Upon  the  exterior  rail,  and  are  more  liable  to  accident  than  slow  trains. 

t  It  has  been  the  practice  of  the  author  to  add  one  fourth  of  an  inch  to  the  computed  elevation. 


164  USEFUL      FORMULAE. 

fore  wheel.  It  may,  however,  be  shown,  from  various  considerations, 
that  if  either  of  the  hind  wheels  produces  friction,  it  is  rather  the 
exterior  one ;  indeed,  we  may  suppose  that  motion  is  communicated 
to  the  hind  wheels  by  a  force  which  acts  precisely  in  the  same 
direction  as  if  they  were  moved  by  animal  power,  the  direction 
being  nearly  a  tangent  to  the  curve.  This  being  admitted,  the 
flanges  of  the  two  exterior  wheels  sustain  all  the  friction  occasioned 
by  curvature.  It  may  be  observed,  however,  that  when  the  distance 
between  the  fore  and  the  hind  wheels  is  comparatively  very  great, 
the  direction  of  the  force  moving  the  hind  wheels  will  vary  con 
siderably  from  the  tangent,  and  consequently  the  friction  will  be 
diminished."  ° 

*  Although  we  agree  with  Mr.  Thompson  in  the  main,  we  do  not  fully  agree  with  his  concluding 
remarks.  Mr.  Thompson  says,  "  It  may,  however,  be  shown,  from  various  considerations,  that  if 
either  of  the  hind  wheels  produces  friction,  it  is  rather  the  exterior  one  ;  indeed,  we  may  suppose  that 
motion  is  communicated  to  the  hind  wheels  by  a  force  which  acts  precisely  in  the  same  direction  as 
if  they  were  moved  by  animal  power,  the  direction  being  nearly  a  tangent  to  the  curve.  This  being 
admitted,  the  flanges  of  the  two  exterior  wheels  sustain  all  the  friction  occasioned  by  the  curvature. 
It  may  be  further  observed,  however,  that  when  the  distance  between  the  fore  and  hind  wheels  is 
comparatively  very  great,  the  direction  of  the  force  moving  the  hind  wheels  will  vary  considerably 
from  the  tangent,  and  consequently  the  friction  will  be  diminished." 

The  reasoning  of  Mr.  Thompson,  doubtless  was  applicable  to  cars  sustained  npon  two  axles  and 
four  wheels  only,  one  axle  being  situated  near  the  forward  end  of  the  car,  and  the  other  near  the  back 
or  hind  end.  Now,  if  the  car  be  short,  the  axles  must  of  course  be  near  each  other  •,  in  this  condition, 
the  flanges  of  both  the  forward  and  hind  wheels  may  grind  the  exterior  rail,  the  forward  wheels  of 
course  grinding  much  the  hardest.  A  distance,  however,  between  the  axles  can  be  readily  ascertained 
which  will  relieve  the  hind  wheels  from  the  friction  of  the  flanges  against  either  the  interior  or 
exterior  rail  •,  then,  expanding  the  distance  between  the  axles,  the  flanges  of  the  hind  wheel  will  begin 
to  grind  against  the  interior  rail,  and  the  greater  the  distances  between  the  axles  the  greater  will  be 
the  friction.  I  would  observe,  however,  that  the  forward  wheel  flange  will,  under  all  distances 
between  the  axles,  grind  upon  the  exterior  rail,  and  will  grind  more  and  more  severely  in  proportion 
as  the  distance  between  them  increases.  These  notions  are  based  upon  the  condition  that  the  axles 
are  firmly  and  permanently  secured  to  the  car,  and  at  right  angles  with  its  frame.  We  might  easily 
demonstrate  the  position  we  have  here  taken  by  diagrams  if  it  were  thought  necessary,  but  the  change 
produced  by  the  adaptation  of  what  we  term  four-wheel  trucks  to  our  long  car  bodies,  which  permits 


ELEVATING      OUTSIDE      RAIL      FOR      CURVE.         165 

We  have  thus  copied  Mr.  Thompson's  article,  with  his  remarks, 
omitting  only  the  portion  relating  to  Mr.  Long's  formula. 

EXAMPLE  OF  COMPUTATION.  Assuming  a  radius,  K  =  4000  feet ; 
and  a  velocity  of  35  miles  per  hour ;  35  miles  per  hour  =  V  = 
51  *33  feet  per  second  of  time;  the  width  between  the  rails  being  «s 
d  =  4  •  7  feet. 

Formula  E  =±  -^— 

R  ±=  4000  feet  co.  ar.        log.  =  6-3979400 

g  =      32-2  co.  ar.        log.  =  8'4921441 

V2  =      51-33°  log.  =  3-4207990 

d  =        4-7  log.  =  0-6720979 


E     =s        0-09615  log.  =  8-9829810 

Again,  for  the  purpose  of  showing  the  changes  of  E,  consequent 
upon  the  changes  of  E,  we  assume  E,  =*  2000  feet;  the  other 
expressions  remaining  the  same.  Thus, 

R  =  2000  feet  co.  ar.        log.  =  6-6989700 

g  =      32-2  co.  ar.        log.  =  8-4921441 

Vs  =      51-33S  log.  =  3-4207990 

d  =        4-7  log.  =  0-6720979 


E     =        0-1923  feet  log.  =  9-2=40110 

We  believe  we  have  in  the  foregoing  pages  examined  every 
distinct  species  of  curve  that  enters  into  the  construction  of  a 
railroad.  We  were  aware,  as  we  proceeded  in  our  investigations, 

the  arrangement  of  the  axles  to  a  very  near  approximation  with  the  radii  of  the  curve,  by  the  force 
of  the  wheel  flanges  against  the  exterior  rail,  seems  to  render  such  an  undertaking  unnecessary, 
particularly  as  the  railroad  companies  appear  to  be  universally  adopting  them.  Butj  it  is  not  our 
intention  to  discuss  generally  the  principles  which  should  govern  the  construction  of  cdrs.  Having 
concluded  to  adopt  the  formula  of  Mr.  Thompson,  we  thought,  in  justice  to  him,  we  Were  bound  to 
copy  his  remarks.  We  would  only  further  mention,  respecting  Mr.  Thompson's  remarks,  that  we 
cannot  discover  any  material  difference  in  the  action  of  the  flanges  of  the  wheels  upon  the  curved 
rails,  whether  the  car  receives  its  motion  from  a  force  pulling  in  front  or  pushing  behind. 


166  USEFUL      FORMULA. 

that  many  modifications  of  the  formula  we  have  deduced  would 
frequently  be  called  for ;  but,  as  we  have  before  stated,  it  has  not 
been  our  intention  to  exhaust  the  subject,  but  merely  to  give  a 
formula  for  the  most  prominent  of  each  class,  or  rather  for  that 
class  which  most  frequently  present  themselves  to  the  engineer 
while  engaged  in  construction. 

We  contemplated,  when  we  commenced  our  work,  closing  our 
paper  here;  but  it  has  occurred  to  us,  that  the  inexperienced 
engineer  might  feel  the  want  of  some  convenient  plan  or  system 
of  computing  the  cubic  contents  of  excavations  and  embankments. 
For  the  purpose  of  supplying  those  wants,  we  add  the  following. 

An  investigation  of  formula  for  the  computation  of  the  cubic 
contents  of  earth,  excavations,  embankments,  masonry,  etc.,  in 
constructing  railroads. 

(59)  An  article  quoted  from  Sillimaris  Journal,  by  Professor 
Eaton,  states,  in  effect,  as  follows ;  that  whereas  the  sections  into 
which  the  engineer  would  divide  the  excavations  upon  a  railroad, 
readily  admit  of  being  subdivided  into  pyramids,  wedges,  and  paral 
lelepipeds;  therefore,  if  you  add  the  area  of  both  ends  of  the  sec 
tion,  to  four  times  its  middle  area,  divide  the  sum  by  six,  and 
multiply  the  quotient  by  the  length  of  the  section,  the  product 
will  give  its  solid  contents. 

This  problem  can  be  readily  demonstrated  to  be  strictly  correct, 
provided  the  sides  of  the  section  are  perfect  planes.  It  is  the 
constant  endeavor  of  every  skilful  engineer  so  to  arrange  the 
sections  that,  were  the  irregularities  of  the  earth  to  be  pared  down, 


MEASURING      SOLIDS      AND      SUPERFICIES.         167 

so  as  to  produce  regular  planes  between  the  points  which  he  has 
chosen  to  take  his  levels  at,  the  solid  contents  of  the  earth  con 
tained  in  the  section  would  he  just  sufficient  to  fill  the  hollows, 
and  make  the  surfaces  planes  between  them. 

(60)  In    Professor    Eaton's   enunciation   of    the   rule   above 
quoted,  I  did  not  discover  any  method  of  ascertaining  the  middle 
area  of  the  section ;  it  being  evident  that  an  arithmetical  mean  of 
the  areas  of  the  ends  of  the  section  would  not  uniformly  produce 
the  desired  result.     To  supply  this  deficiency  in  the  formula  is  the 
main  object  of  the  present  paper ;  but,  as  the  original  formula  may 
not  be  within  the  reach  of  every  individual  who  may  feel  interested 
in  seeing  an  investigation  and  demonstration  of  it,  we  have  thought 
a  brief  investigation  might  not  be  out  of  place ;  besides,  it  will 
aid  much  in  rendering  the  subject  more  plain  and  intelligible.     I 
shall,  however,  take  it  for  granted  that  the  interested  reader  will 
know  enough  of  geometry  to  be  familiar  with  the  common  formulae  for 
measuring  the  solid  contents  of  pyramids,  parallelepipeds,  wedges, 
and  surfaces  of  the  cross  sections  of  the  excavations  of  a  railroad ; 
we  shall,  therefore,  only  allude  to  the  most  common  formulae  for 
measuring   solids   and    superficies,    as   we   may   have   occasion   to 
compare  them  with  the  formula  to  be  deduced. 

Commencing  with  the  pyramid. 

(61)  According  to  the  rule,  we  have  to  add  the  area  of  the 
base  of  the  pyramid  to  four  times  the  area  of  its  middle  section, 
(taken  parallel  with  said  base,)  divide  the  sum  by  six,  and  multiply 
the  quotient  by  the  height  of  the  pyramid ;  the  product  will  give 
its  solid  contents. 


[Fia.     26(i) 

THE    PYRAMID. 


G.     26(2).] 

THE    WEDGE. 


PYRAMID      AND      WEDGE.  169 

For  the  purpose  of  proof,  or  demonstration,  let  us  suppose  a 
pyramid  of  four  sides,  Fig.  1,  the  angles  of  its  base  being  right 
angles,  and  the  lines  circumscribing  said  base  of  equal  length, 
which  length  we  represent  by  a;  it  is  evident  that  the  lines  which 
circumscribe  the  middle  section  of  said  pyramid  (taken  parallel  with 
said  base)  will  be  equal  to  J  a;  if  then  we  represent  the  height  by 
h,  we  have  for  the  area  of  the  base  a  X  a  =  a2;  for  the  area  of 
the  middle  section  |aXi^==i«2;  then,  by  the  rule,  (  fl3  +  *«34  j  h 
=  A?aJL  =  ^  a2  h  =  the  solid  contents  ;  which  corresponds  exactly 


with  the  formula  in  common  use  (1.) 

This  equation  shows  the  area  of  the  middle  section  of  a  pyramid 
to  be  one  fourth  of  the  area  of  the  base  ;  and  this  proposition  is 
universal  and  equally  correct  in  every  species  of  pyramid,  whether 
it  be  three,  four,  or  many-sided,  regular  or  irregular. 

Secondly,  the  wedge. 

(62)  We  next  apply  the  rule  to  the  measurement  of  the  wedge. 
Let  us  now  suppose  a  regular  or  symmetrical  wedge,  with  a  base 
circumscribed  by  lines  of  equal  length,  which  we  represent  by  a;  it 
is  obvious  that  of  the  lines  which  circumscribe  the  middle  section  of 
the  wedge,  (taken  parallel  with  the  base,)  two  of  them  will  equal 
a;  and  the  other  two  will  equal  J  a.  Then,  making,  as  "before,  h 
equal  the  height,  or  length  ;  we  have,  by  the  rule, 

For  the  area  of  the  base     a  X  a  =  az  ; 
For  middle  section      a  X  \  a  =  J  <£- 

Then,  (  a2  +/  fl2  4  )h=  *^h-  =  i  a2  h  =*  the  solid  contents  ; 
which  corresponds  with  the  formula  in  common  use  for  determining 

the  solid  contents  of  the  wedge  (2.) 
x 


170  USEFUL     FORMULA. 

From  this  equation  we  learn,  that  the  area  of  the  middle  section 
is  equal  to  one  half  the  area  of  the  base.  By  cutting  the  wedge 
into  pyramids  we  may  compute  its  solid  contents  in  a  way  some 
what  different,  with  the  same  formula. 

Let  A  B  C  D,  in  Figure  2,  represent  the  base  end  of  the  wedge; 
and  a  b  the  edge  or  sharp  end ;  cut  the  wedge  diagonally  through 
the  plane  6  B  C.  The  wedge  is  thus  divided  into  two  pyramids; 
A  B  C  D  b,  and  a  b  B  C ;  the  pyramid  A  B  C  D  b  being  a  four- 
sided  one ;  and  the  pyramid  a  b  B  C  being  a  triangular  or  three- 
sided  one.  We  may  further  cut  the  four-sided  pyramid  in  the 
plane  b  B  D,  which  divides  that  pyramid  into  two  three-sided  pyra 
mids.  The  wedge  will  then  consist  of  three  triangular  pyramids ; 
but,  as  the  same  rule  for  determining  the  cubes  applies  to  three, 
four,  and  many-sided  pyramids,  we  shall,  in  our  further  investi 
gations  of  the  mensuration  of  the  wedge,  only  use  the  four-sided 
pyramid,  in  connection  with  the  blind  pyramid  a  b  B  C ;  (this 
pyramid  is  so  named  because  it  presents  no  area  in  either  of  the 
surfaces  of  the  cross  section  of  a  cut  in  a  railroad  excavation.) 
Pyramids  of  this  character  enter  into  the  calculations  of  nearly 
every  cross  section,  and  my  principal  object  in  introducing  it  in  the 
wedge,  is,  for  the  purpose  of  testing  the  method  of  computing  its 
solid  contents* 

If  we  now  compare  the  solid  contents  of  the  four-sided  pyramid, 
it  will  be  observed  that  we  make  use  of  the  whole  area  of  the  base 
of  the  wedge,  and  add  thereto  four  times  the  quarter  area  of  the 
base ;  which  quarter  is  equal  to  one  half  the  area  of  the  middle 
section  of  the  wedge  ;  this  sunl,  divided  by  six,  and  the  quotient 
multiplied  by  the  length  of  the  wedge,  gives  the  solid  contents  of 


PYRAMID      AND      WEDGE.  171 

the  pyramid.  Then,  to  complete  the  measurement  of  the  wedge, 
according  to  the  rule  ;  having  divided  the  area  of  the  base,  plus 
four  times  one  fourth  of  the  area  of  the  base,  etc. ;  as  explained 
before,  the  remainder  of  the  area  of  the  middle  section  will  of  course 
be  equal  to  one  fourth  of  the  area  of  the  base ;  which,  multiplied  by 
four,  its  product  divided  by  six,  and  the  quotient  multiplied  by  the 
length  of  the  wedge,  the  operation  will  be  complete ;  (and  have  been 
performed  in  a  different  method,)  the  result  being  the  same  as  in 
equation  (2.) 

I  would  however  mention,  that  we  found  by  equation  (2)  that 
the  middle  area  of  the  wedge  was  equal  to  half  the  area  of  the 
base ;  and,  in  measuring  the  four-sided  pyramid,  the  middle  area  of 
which  is  equal  to  one  quarter  of  the  area  of  the  base,  of  course  it 
is  equal  to  one  half  of  the  middle  area  of  the  wedge,  leaving  the 
other  half  for  the  middle  area  of  the  blind  pyramid. 

To  elucidate  this,  let  us  introduce  the  calculations. 

Using  the  former  notations,  we  have,  in  the  mensuration  of  the 
four-sided  pyramid,  for  the  base  a  X  a  =  a2,  the  middle  area  J  a 
X  i  a  =  k  a* ;  then,  (J*+£*±)  .  h  =  -2  ^  A  (3.) 

In  the  mensuration  of  the  blind  pyramid  a  b  B  C,  we  have  for 
the  middle  area,  (a  b  being  equal  to  a,  and  B  C  being  also  equal 
to  a)  \  a  X  i  a  =  J  az ;  and  (_*  <£A.)  .  h  =  -«'.*-  (4.) 

If  we  now  add  the  results  of  the  above  equations,  (3  and  4,) 
their  sum  will  be  equal  to  the  contents  of  the  wedge,  as  found  in 
equation  (2  ;)  thus  proving  that  the  mensuration  of  the  blind  pyra 
mid  is  exact,  according  to  the  rule.  As  a  further  proof,  we  may 


[FiG.     27.] 


IN   WHICH  THE  TWO   SEPARATE  DIAGRAMS  DEFERRED  TO  BY  THE  TEXT  MAY  BE  TRACED. 


EXCAVATIONS       AND      EMBANKMENTS.  178 

DOW  measure  the  blind  pyramid  in  this  manner.  Making  a  B  C 
the  base,  and  a  b  the  height,  which  we  shall  denote  by  h;  then, 
according  to  rule,  and  using  the  same  notation,  we  have 

/  i  a  h  ± -a La_xjr _*)_ii  )  .  a  =    a\h     (5;) 

the  same  as  in  equation  (4.) 

We  might  produce  a  great  variety  of  proofs  to  show  the  accuracy 
of  the  rule.  Thus,  if  we  should  endeavor  to  measure  a  four-sided 
square  parallelepiped  by  the  rule,  the  simplest  manner  of  proceed 
ing  will  be  to  divide  it  into  two  wedges,  and  then  apply  the  rule ; 
or,  we  may  cut  off  four  pyramids,  leaving  a  large  blind  pyramid, 
which,  in  order  to  determine  its  contents,  will  require  that  we  should 
determine  the  length  of  the  diagonals,  (this  method  will  be  exact 
only  when  the  diagonals  are  at  right  angles),  and  then  multiplied 
into  each  other,  will  give  four  times  the  middle  area  required ;  this, 
divided  by  six,  and  multiplied  by  the  length,  will  give  the  solid 
contents  of  the  blind  pyramid. 

But,  this  last  method  is  somewhat  complicated,  inasmuch  as  we 
should  be  obliged  to  find  the  diagonals  by  extracting  their  square 
roots  from  the  sum  of  the  squares  of  the  other  sides ;  or,  we  may 
find  the  diagonal  by  trigonometry.  I  shall,  therefore,  only  give  an 
example  of  determining  the  cubic  contents  of  the  parallelopiped,  by 
dividing,  first,  into  two  wedges.  This  example  we  give  merely  as 
an  illustration  of  the  method, 

Retaining  the  former  notation,  we  have  for  the  measurement  of 
the  apparent  pyramids,  the  bases  and  four  times  the  middle  area  of 
each  pyramid,  equal  to  twice  the  area  of  the  bases.  See  equation 
(1,)  which,  for  both  pyramids,  is  equal  to  4  «2,  and  four  times  the 
middle  area  of  one  of  the  blind  pyramids,  will  be  a  X  ct  =  a*  each 


174  USEFUL      FORMULAE. 

there  being  two  of  them,  we  have,  therefore,  4  a2  -f-  2  az  =  6  az, 
and  6  a\  h  =  a2  h  ;  the  same  result  as  by  the  ordinary  method  of 
determining  the  cubic  contents  of  a  parallelepiped. 

Of  course  the  above  is  not  the  most  convenient  method,  but  will 
serve  to  show  the  application  of  the  rule  to  the  measurement  of  the 
frustrum  of  a  pyramid. 

The  rule  is  also  peculiarly  applicable  to  the  measurement  of 
wedges  in  which  one  end  is  wider  than  the  other,  and  to  almost 
every  figure  imaginable  which  is  bounded  by  right  lines  and  plane 
surfaces. 

We  will  now  give  the  calculations  of  a  few  imaginary  figures  of 
different  forms,  after  the  manner  of  our  practice,  to  determine  the 
cubic  contents  of  excavations,  embankments,  masonry,  etc. 

(63)  Let  A  B  C  I L  K  represent  one  end  of  the  supposed  section, 
which  we  will  denominate  No.  1,  (in  Fig.  27,)  and  D  E  F  G  M  H,  the 
other  end,  which  we  denominate  No.  2.  After  preparing  diagrams 
of  the  ends  of  the  sections,  and  marking  the  heights  of  the  points 
A  B  C  of  diagram  No.  1,  and  of  D  E  F  in  No.  2,  we  then  divide  the 
diagrams  into  figures  which  we  shall  now  proceed  to  describe. 

Firstly,  we  divide  No.  1  by  the  perpendicular  line  L  B,  the  point 
L  representing  the  centre  of  the  road  bed. 

Secondly,  draw  the  line  A  c  parallel  to  the  base,  or  the  road  bed, 
K  I  ;  which  is  always  level. 

Thirdly,  draw  the  line  B  a  parallel  to  the  line  K  I ;  then  the 
diagram  will  be  divided  into  the  triangles  A  B  c  and  B  C  a,  and  the 


EXCAVATIONS      AND      EMBANKMENTS.  175 

trapczoids  A  K  L  c  and  B  L  I  a.  We  now  propose  for  the  dimensions 
that  K  L  and  L  I  shall  each  be  9  feet  long,  and  that  the  height  of 
A  he  4  feet,  the  height  of  B  8  feet,  the  height  of  C  12  feet.  In 
order  to  determine  the  areas  of  the  above  mentioned  trapezoids  and 
triangles,  we  first  determine  the  length  of  the  lines  A  c  and  B  a; 
the  length  of  A  c  being  equal  to  the  height  of  A,  plus  one  half  the 
height  of  A  -j-  K  L,  upon  the  supposition  that  the  slopes  of  the 
cuttings  are  as  three  to  two.  Thus, 

The  height  of  A =    4  feet 

&  "  " =2 

K  L  =9 

Length  of  A  c =15  feet 

which  we  mark  upon  diagram  No.  1. 

To  find  the  length  of  B  a,  we  have  the  height  of  B  L  +  }  the 
height  of  B  L  +  L  I  =  B  a.     Thus, 

The  height  of  B  L  =8  feet 

*  "  " =    4 

Length  of  L  I  =9 

Length  of  B  a •         .        =21  feet 

which  we  mark  upon  the  diagram  also. 

Having  prepared  the  diagram,  we  proceed  to  determine  the  area 
of  the  trapezoid  A  K  L  c.     We  have  found 

Ac  =  15  feet 

K  L  =9 

AC  +  KL 2)24 

Mean,  or  }  (A  c  +  K  L) =12 

Height  of  A  above  K  =  L  c =4 

Area  =  48  feet 


USEFUL      FORMULA. 

To  find  the  area  of  the  trapezoid  a  B  L  I,  we  have  found 

a  B                 =  21  feet 

LI =  9 

a  B  +  L  I)  =  2)30 

k  (a  B  +  L  1) =  15 

Height  of  B  —  8 


Area =          120  feet 

To  find  the  area  of  the  triangle  A  B  c,  we  have  found 

Ac  =     15  feet 

i  difference  of  heights  B  and  A  =      2 

Area     .  =30  feet 

To  find  the  area  of  the  triangle  a  B  C,  we  have  found 

B  a =    21  feet 

£  difference  of  heights  of  C  and  B      .         .         .         .        =      2 

Area =42  feet 

Having  determined  the  areas  of  each  figure  composing  diagram 
No.  1,  and  marked  the  same  upon  it,  we  then  proceed  to  divide 
diagram  No.  2  into  triangles  and  trapezoids,  and  compute  their 
areas  in  a  manner  similar  in  principle  to  that  adopted  in  No.  1. 

Assuming  the  height  of  F  ==  8  feet;  the  height  of  E  =  13 -6 
feet ;  the  height  of  D  ==  10  feet ;  and  G  M  and  M  H  ==  9  feet 
in  length  each.  Firstly,  we  have,  in  the  triangle  F  g  E, 

The  length  of  the  line  F  #,  equal  the  height  of  F     .        =8  feet 

+  half  the  height  of  F =4 

+  GM =9 

Wherefore,  F  5- =    21  feet 

We  then  have  half  the  difference  in  heights  of  g  and  E  =  half 
the  difference  of  F  and  E.  Thus, 


EXCAVATIONS      AND      EMBANKMENTS.  177 

The  height  of  F =    8      feet 

«        «  E  =  13.6 


Difference 2^5.6 

Half  difference  of  g  E =    2-8  feet 

We  now  find  the  area  of  the  triangle  F#E  =  21X2'8  = 
58-8  feet. 

Secondly.  In  the  triangle  D  E  /we  have 

The  length  of  the  line  D/  equal  the  height  of  D       .  =10  feet 

+  "        "        }  D        .  =5 

+  M  H       .        .        .  =9 

Wherefore  /  D =    24  feet 

Then  we  have  half  the  difference  in  height  of  /  and  E  equal  to 
half  the  difference  of  D  and  E  ;  thus, 

The  height  of  D =    10-00  feet 

"        "            E =    13-60 

Difference 2)  3-60 

Half  difference 


We  now  find  the  area  of  the  triangle  DE/=24Xl'8  = 
43-2  feet. 

Thirdly.  To  find  the  area  of  the  trapezoid  F  Gr  M  g ;  we  found 
in  the  foregoing, 

F  g- =    21-00  feet 

G  M  =      9-00 


FS-  +  GM         2)30-00 

Mean  length  =    15-00 

Then,  M  G  =  height  of  F =      8-00 


Area =  120*00  feet 

Y 


[FiG.     28.] 


EXCAVATIONS      AND      EMBANKMENTS.  179 

Fourthly.  Then,  to  find  .the  area  of  the  trapezoid  D  H  M/;  we 
find  in  the  foregoing, 

D/ =    24-00  feet 

M  II  =      9-00 


D/+  M  H  .         .         .     "    .         .         .         .  2W 


00 


Mean  length  =    16-5 

Then,  M /=  height  of  D =    lO'OO 


Area =  165-00  feet 

Having  ascertained  the  several  areas  of  the  divisions  of  the  cross 
sections,  and  marked  the  same  upon  the  diagram,  our  next  operation 
will  be  to  ascertain  the  cuhic  contents  of  that  portion  of  the  half 
section  lying  above  the  lines  F  g,  g  c,  c  A,  and  A  F. 

(64)  It  will  be  evident,  from  an  inspection  of  the  drawing,  that 
the  upper  portion  of  the  half  section  may  be  divided  into  two  appar 
ent,  and  one  blind  pyramid.  This  solid  admits  of  two  distinct  methods, 
or  plans  of  divisions,  so  that,  if  the  surface  is  not  twisting,  it  matters 
not  which  of  the  plans  is  adopted.  For  the  purpose  of  a  test,  we 
will  consider  the  divisions  under  both  aspects  ;  firstly,  as  represented 
in  the  drawing  of  the  section,  (viz.,  Fig.  27 ;)  and  secondly,  as  in 
Fig.  28.  Whatever  difference  there  may  be,  if  any,  will  be  seen  in  the 
different  dimensions  the  blind  pyramid  will  assume ;  hence  we  will 
confine  ourselves  to  the  comparison  of  the  contents  of  this  solid. 

Firstly,  we  have  A  c  X  g  E  =  4  times  the  centre  area  of  the 
blind  pyramid ;  and,  in  the  second  form,  we  have  F  g  X  c  B  =  4 
times  the  centre  area,  as  above. 

FIRST  FOEM.  SECOND  FORM. 


Ac.         .         .  =    15-       feet 

g  E  .         .         .        =5-6 


9-0 
75- 
4  times  middle  area      .        =84-00 


F  g       .        .         .         .        =    21-00  feet 
c  B       .  =  ~'4'00 


4  times  middle  area  =A 84-00 


.     29.] 


EXCAVATIONS      AND      EMBANKMENTS.  181 

It  thus  appears  that  the  top  surface  is  a  perfect  plane :  had  it 
been  warped,  or  twisted,  the  middle  areas  of  the  blind  pyramid 
under  both  aspects  would  not  have  been  alike. 

We  have  now,  in  the  remainder  of  the  first  half  section,  the  solid 
A  K  L  eg  F  G  M.  The  lines  F  g,  G  M,  K  L,  A  c,  being  parallel, 
the  solid  admits  of  a  division  into  two  wedges ;  and  these  wedges, 
as  we  have  already  shown,  may  be  divided  into  one  apparent  and 
one  blind  pyramid  each  ;  the  area  F  g  M  G  being  the  base  of  one  of 
the  apparent  pyramids,  and  the  area  A  c  L  K  being  the  base  of  the 
other.  The  surfaces  of  this  solid  being  perfect  planes,  it  matters 
not  in  what  manner  we  form  the  wedges,  and  cut  from  them  the 
blind  pyramids,  as  their  combined  measurements  will  be  the  same. 
For  example,  if  the  wedges  be  formed  by  cutting  the  solid  through 
the  plane  A  c  M  G,  four  times  the  middle  area  of  one  of  the  blind 
pyramids  will  be  equal  to  M  g  X  A  c,  and  four  times  the  middle 
area  of  the  other  will  be  equal  to  L  c  X  G  M.  Or,  we  may  form 
the  wedges  by  cutting  the  solid  through  the  plane  K  L  g  F  ;  then, 
four  times  the  middle  area  of  one  of  the  blind  pyramids  will  be 
equal  to  L  c  X  F  g,  and  the  other  will  be  equal  to  g  M  X  K  L. 
Before  we  proceed  further  with  our  investigation,  we  will  compare 
the  contents  of  the  blind  pyramids  as  ascertained  by  both  methods 
of  division. 

Firstly.  We  have  A  c  X  M  g  =  4  times  the  middle  area  of  one 
of  the  blind  pyramids. 

Ac =15  feet 

MS- =8 

4  times  the  middle  area, =120  feet 


182  USEFUL      FORMULA. 

Again,  we  have  L  c  X  G  M  =  4  times  the  middle  area  of  the 
other  blind  pyramid.     Thus, 

Lc =      4  feet 

G  M =9 

4  times  the  middle  area =36 

Then  add  the  area  found  above  =  120 

Gives  the  sum  of  the  middle  areas  of  both         .         .  =  156 

Secondly.  We  have  L  c  X  F  g  =  4  times  the  middle  area  of  one 
of  the  blind  pyramids.     Thus, 

Lc =4  feet 

F  g =21 

4  times  the  area =84  feet 

Again,  we  have  M  g  X  K  L  =  4  times  the  middle  area  of  the 
other  blind  pyramid.     Thus, 

M  g =      8  feet 

KL =9 

4  times  the  area =    72 

Add  the  area  found  above =84 

Sum  of  middle  areas  of  both =156  feet 

Hence  we  see  that  the  computations  prove  that  the  final  result 
will  be  the  same  under  both  methods  of  computation. 

We  will  now  sum  up  the  measurement  of  the  supposed  divisions 
of  the  first  half  section. 


EXCAVATIONS      AND      EMBANKMENTS.  183 

1st.    We  have  the  area  of  the  base  of  the  triangle  F  g  E  A  =  58 -8  feet,  and  +  4  times 

its  middle  area  =  58-8  ;*    (See  Fig.  27) =    117 -6  feet 

2d.    We  have  the  area  of  the  base  of  the  pyramid  A  B  c  E  =  30  feet,  and  +  4  times  its 

middle  area  =  30 =      60*0 

3d.    We  have  4  times  middle  area  of  blind  pyramid  A  c  E  g-  .         .         .         .        =      84*0 

4th.     We  have  the  area  of  the  base  of  the  pyramid  F  G  M  g  A  =  120  feet,  and  +  4 

times  the  middle  area  =  120 =240-0 

5th.    We  have   the   area  of  the  base  of  the  pyramid  A  c  L  K  g  =  48  feet,  and  -f  4 

times  the  middle  area  =  48 =      96-0 

6th.    We  have  the  sum  of  the  middle  areas  of  the  remaining  two  blind  pyramids  =    156*0 


753-6 
7th.    Taking  the  length  of  the  section  =  100  feet 100 


Dividing  by 6^75360-0 

Solid  contents  =  1256.0 

(65)  It  may  not  be  amiss  here  to  remark,  that  we  always,  when 
the  nature  of  the  case  will  admit,  divide  each  section  into  two  parts 
by  a  plane  passing  through  the  vertical  lines  M  E  and  B  L,  and 
compute  each  portion  separately,  (see  Fig.  27.)  The  reason  for  this  is, 
that  a  level  is  always  taken  over  the  centre  of  the  road  bed  in  every 
cross  section,  and,  in  a  majority  of  the  cases  which  occur,  there  are 
only  two  other  levels  taken  ;  viz.,  one  at  the  right,  and  one  at  the 
left  hand  slope  stakes ;  and  whenever  it  becomes  necessary  to  take 
other  levels  at  the  right  or  left  of  the  centre,  it  will  still  be  con 
venient  to  preserve  the  centre  division.  And  it  will  be  apparent 
when  we  have  completed  our  computations,  that  our  system  of 
dividing  the  cross  sections  into  triangles  and  trapezoids,  is  pecu- 


*  In  the  early  part  of  this  discussion,  we  proved  that  the  middle  area  of  a  pyramid,  taken  parallel 
to  its  base,  was  equal  to  one  fourth  of  the  area  of  its  base  ;  hence,  4  times  the  middle  area  will  be 
equal  to  that  of  the  base. 


184  USEFUL      FORMULA. 

liarly  adapted  to  this  method  of  computing  cubic  contents,  and 
affords  a  very  convenient,  as  well  as  an  accurate  method  of  ascer 
taining  the  area  of  said  cross  section. 

Before  we  proceed  to  the  computation  of  the  cubic  contents  of  the 
remaining,  or  second  half  section,  we  would  remark  that  the  portion 
of  the  second  half  section  which  lies  above  the  plane  passing 
through  B/D  a,  having  its  upper  surface  much  twisted  or  warped, 
admits  of  two  forms  of  division,  which  will  be  found  by  computation 
to  give  different  results  ;  one  of  which  will  be  applicable  to  one 
form  of  surface,  and  the  other  to  another.  And  that  the  portion  of 
the  half  section  lying  below  the  plane  B  /  D  a  having  all  of  its  sur 
faces  perfect  planes,  the  computation  will  give  correct,  and  of  course 
like  results,  from  whichever  of  the  forms  the  divisions  may  take. 

We  now  proceed  to  the  examination  of  the  upper  portion  of  said 
half  section. 

Firstly.  If  we  suppose  the  upper  surface  to  have  this  form,  viz., 
that  of  a  plane  through  DEB,  and  intersecting  in  the  line  D  B,  a 
plane  passing  through  DBG,  the  solid  will  then  contain  only  two 
apparent  pyramids,  with  no  blind  pyramid;  viz.,  the  apparent 
pyramids  D  f  E  B  and  B  a  C  D  ;  the  measurement  of  which  is  as 
folio  AYS  : 

For  the  area  of  base  of  D  E  /  B  =  43-2  feet,  and  -f  4  times  middle  area  =  43-2         =      86-4  feet 
a  B  C  D  =  42        «      «     _}-  4     «          «          «     =  42  =      84'0 


Dividing  by  e)l70'4 

28-4 
Multiplying  by  the  length  of  section  100 


Solid  contents  2840-0 


EXCAVATIONS      AND      EMBANKMENTS.  185 

Secondly,  if  we  suppose  the  upper  surface  to  have  the  following 
form  ;  viz.,  that  of  a  plane  passing  through  the  points  B  C  E,  and 
intersecting  in  the  line  C  E,  a  plane  passing  through  the  points 
C  D  E  ;  the  solid  will  then  contain  two  apparent  pyramids  and  two 
blind  pyramids,  viz.,  the  apparent  pyramids  B  a  C  E  and  D  E  /C, 
and  the  blind  pyramids  C  a  f  D  and  E  /  a  B  ;  to  measure  which 
we  have 

For  the  area  of  base  of  D  E/  C  =  43 -2  feet,  and  +  4  times  middle  area  =  43-2  =    86 -4  feet 

"            "             "          aBCE  =  42        "            +4      "          "         "     =  42  =    84'0 

"     4  times  middle  area  of  blind  pyramid  C  a/D  =  C  a  X/D  =4  X  24       ,         .  =    96-0 

«            «          «          "            "            "         aBE/=aB  X  E/=21  X  3-6    ,         .  =    75-6 

Dividing  by 6^342-0 

57-0 

Multiplying;  by  length  of  section 100 


5700-0 


We  thus  find  the  upper  portion  of  the  second  half  section  under 
consideration  computed. 

For  the  1st  form  of  surface  gives  solid  contents =  2840  feet 

u      u    2ll      «  »  »        u          u  .        =  5700 


Difference  in  cubic  feet 27^2860 

"         "      "      yards =106 

The  wide  difference  in  the  results  shows  the  necessity  of  noting 
while  in  the  field  the  form  of  the  twisted  surface,  that  the  proper 
method  of  computation  may  be  applied.  But  it  is  not  supposed 
that  surfaces  like  those  we  have  been  considering  will  very  fre 
quently  occur  in  practice,  as  the  engineer  would  be  likely  to  divide 
z 


186  USEFUL      FORMULAE. 

the  section  into  two  or  three,  which  would  have  a  tendency  to  lessen 
the  differences  much.  But,  to  repeat,  if  the  form  of  a  surface  he 
noticed  when  the  field  work  is  performed,  so  that  the  proper  form  of 
dividing  the  cross  section  may  he  applied,  large  sections  may 
frequently  he  computed  with  as  good  a  degree  of  accuracy,  or  even 
"better,  than  smaller  sections  without  such  notice. 

To  complete  the  computation  of  the  cuhic  contents  of  the  figure, 
we  have 

The  double  area  of  the  base  of  pyramid  H  M/  D  a =    330  feet 

Also    "          "         "          "  "       a  B  L  I  H =240 

Then,  4  times  the  middle  area  of  theblind  pyramid  =/M  X  B  a  =  21  X  10      .         .  =210 

And    4     "        "        "          "        "          "  "        =  B  L  X  M  II  =  9  X  8        .         .  =72 

Dividing  by 6)852 


142 
Multiplying  by 100 


Cubic  contents  of  aBLIHM/D =    14200 

Having  thus  discussed  the  operations  necessary  to  ohtain  the 
cuhic  contents  of  every  portion  of  the  section,  we  now  add  an  ex 
ample  of  summing  up  the  contents  after  the  manner  in  common 
practice.  Before  entering  on  our  work,  we  remark  that  we  contem 
plate  two  summations  ;  the  first  containing  the  computation  of  the 
upper  portion  of  the  second  half  section,  noticed  in  the  foregoing  as 
containing  no  hlind  pyramid  ;  and  the  second  containing  the  com 
putation  of  the  upper  portion  of  said  second  half  section,  noticed 
as  containing  two  hlind  pyramids. 

(66)  Now,  as  we  have  the  areas  computed,  and  marked  upon 
the  diagram  of  the  cross  sections,  as  described  in  the  foregoing,  we 
have  as  follows  : 


EXCAVATIONS      AND      EMBANKMENTS.  187 

FIRST  SUMMATION.  SECOND  SUMMATION. 

No.  1    30-00  feet  taken  twice    =    GO'OO  feet  60 -00  feet 

"    2    48-00     "        "        "         =    96-00  96'00 

"    3    58-8       «        "        "        =  117-60  117-60 

"    4  120-00      "        "         "        =  240-00  240'00   ^ 

Blind  pyramid     "    1    E  g  X  A  c  =  15  X  5'6  —    84-00  84'00 

"    2    g-MxAc  — 15X8      =  120'00  120-00 

"3LcxGM  =  4x9        =    36-00  36'00 

"    5    42-00  feet  taken  twice    =    84-00  84-00 

"    6  120-00      "         "        "        =  240-00  240'00 

"    7    43-2        "         "         "         =    86-4  86-4 

"    8  165-00      "         "         "         =  330-00  330'00 

Blind  pyramid     "    1  =  M/  X  B  a  —  10  X  21  =  210-00  210-00 

"    2  =  B  L  X  M  H  =  8  X  9   =    72-00  Blind  pyramid                         72-00 

Dividing  by 6^1776-00  acX/D  —  4x24      =    96-00 

296-00  E/X  B  A  =  3-6  X  21  =    75'6 


Multiplying  by  length  of  section         ...        100  Dividing  by     .        .      6)1847-6 


27^29600-00  324.6 

Cubic  contents,  according  to  1st  summation  =    1096-30  yds.      Length  of  section  =      100 


1202-22  27  \ 32460 -00 

Difference  ^=      105-92  Cubic  yards       .         .         1202'22 

The  figures  considered  in  the  foregoing  pages  are  those  most 
commonly  met  with  in  railroad  excavations.  We  frequently  meet 
with  modifications,  however,  containing  points  necessary  to  he 
noticed,  between  the  slope  and  centre  stakes  ;  hut  it  is  believed  that 
the  ingenious  engineer  will,  with  a  little  practical  experience,  be 
enabled  so  to  arrange  the  division  of  the  sections,  whatever  may  be 
their  form,  into  pyramids,  so  as  to  admit  of  a  ready  and  satisfactory 
method  of  computing  their  solid  contents. 

Before  we  leave  this  subject,  we  would  remark,  that  the  general 
rule  we  have  been  endeavoring  to  demonstrate  gives  the  measure 
ment  of  the  middle  area  of  the  blind  pyramid  C  «/D  in  the  second 


188  USEFUL      FORMULAE. 

half  section,  a  trifle  large  ;  but  so  near  the  truth,  that  it  has  not 
been  deemed  necessary  to  change  or  modify  the  form  of  division  or 
computation.  In  the  supposed  figure,  named  above,  which  presents 
rather  an  uncommon  case,  the  error  is  something  less  than  a  square 
yard.  If  the  length  of  the  section  had  been  taken  at  some  thirty, 
or  even  fifty  feet,  the  error  would  have  been  much  diminished. 

We  had  thought  of  adding  some  formula  for  computing  the 
cubic  contents  of  what  we  technically  term  borrowing  pits  ;  but,  as  a 
great  majority  of  the  figures  composing  these  pits  possess  forms  or 
solids  so  simple  in  their  character  that  they  will,  at  first  thought, 
suggest  ready  and  appropriate  methods  of  computation,  we  deem 
it  unnecessary  to  further  enlarge  upon  the  subject. 


(T 


YC   13171 


M166992 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


SBH.OII.Y&C1 

v    >7.u 


